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let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
so minus 24/15 and we get it being equal to 16/15 . and so using green 's theorem we were able to find the answer to this integral up here . it 's equal to 16/15 .
what if i find some singularities inside the closed curve , how should i proceed in order to analyse the integral by green 's theorem ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
hopefully you found that useful . i 'll do one more example in the next video .
but according to another video , a conservative field is where the `` density '' does n't change ( fluid example , maybe grant 's video ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
you take the dot product of this with dr , you 're going to get this thing right here . so this expression right here is our p of x y . and this expression right here is our q of xy .
how do you get the functions p ( x , y ) and q ( x , y ) if you are given the region r ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
it would have been a blunder . it 's 8/3 . 8/3 times 1 to the third minus 8/5 times 1 to the fifth , so that 's minus 8/5 .
how did you come up with those boundaries 0 ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
this is what equals the double integral over this region r of the partial of q with respect to x minus the partial of p with respect to y d area . and just as reminder , now this q and p are coming from the components of f in the situation . f would be written as f of xy is equal to p of xy time the i component plus q ...
what if f had a p , q , and r component could greens theorum still be applied ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
and that 's the situation which green 's theorem would apply . so if you were to take a line integral along this path , a closed line integral , maybe we could even specify it like that . you 'll see that in some textbooks -- along the curve c of f dot dr .
in general , how can i write double integral where y is the second integral and x is the first ( inside ) integral and to give the same answer ?
let 's see if we can use our knowledge of green 's theorem to solve some actual line integrals . and actually , before i show an example , i want to make one clarification on green 's theorem . all of the examples that i did is i had a region like this , and the inside of the region was to the left of what we traversed...
and that 's the situation which green 's theorem would apply . so if you were to take a line integral along this path , a closed line integral , maybe we could even specify it like that . you 'll see that in some textbooks -- along the curve c of f dot dr .
this is relating a line integral of vector field to the curl of the vector field , is n't it ?
so we know that when a cerebral artery gets blocked the brain tissue that that vessel serves can die off pretty quickly . the neurons die off without the oxygen that they get from their blood supply , right ? and that 's what a stroke is . well , it turns out that what we see is this interesting pattern in brain tissu...
which is also called necrosis . and this actually happens in only minutes . it only takes a few minutes for the core neurons to die off .
what happens to the dead tissue ?
so we know that when a cerebral artery gets blocked the brain tissue that that vessel serves can die off pretty quickly . the neurons die off without the oxygen that they get from their blood supply , right ? and that 's what a stroke is . well , it turns out that what we see is this interesting pattern in brain tissu...
if for whatever reason blood flow stops to some part of your brain . you develop an ischemic core and ischemic penumbra . neurons in their supporting cells that are in the ischemic core , they 'll die off within a few minutes .
does an ischemic core develop when one experiences a tia ?
so we know that when a cerebral artery gets blocked the brain tissue that that vessel serves can die off pretty quickly . the neurons die off without the oxygen that they get from their blood supply , right ? and that 's what a stroke is . well , it turns out that what we see is this interesting pattern in brain tissu...
so just a quick recap now . if for whatever reason blood flow stops to some part of your brain . you develop an ischemic core and ischemic penumbra . neurons in their supporting cells that are in the ischemic core , they 'll die off within a few minutes .
or is the effect of a tia more like the ischemic penumbra , where the cells experience oxygen and nutrient deprivation but regain function when blood flow is restored ?
so we know that when a cerebral artery gets blocked the brain tissue that that vessel serves can die off pretty quickly . the neurons die off without the oxygen that they get from their blood supply , right ? and that 's what a stroke is . well , it turns out that what we see is this interesting pattern in brain tissu...
if for whatever reason blood flow stops to some part of your brain . you develop an ischemic core and ischemic penumbra . neurons in their supporting cells that are in the ischemic core , they 'll die off within a few minutes .
is there a sharp distinction between the ischemic core and penumbra , or do these two areas sort of 'blend ' into one another ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
so like , there 's a lot of reasons why you might pause in your speech , you might be hesitating or thinking or just whatever the case may be , you can use an ellipsis to show that you 're pausing . - [ david ] right . uh , and usage number two for the ellipsis is that it shows that a portion of quoted material has bee...
does n't that make it clearer , especially if you then put a full stop right after the ellipse and end up with 4 points in a row ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ david ] if you did n't use ellipses , you could just render that however you please , because those were words that were said approximately in that order , right . - [ paige ] yeah , without ellipses or -- - [ david ] without ellipses . - [ paige ] okay .
if i change deleted part of a quote , should i add a space before ellipses ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ paige ] okay . - [ david ] use number one . it indicates a pause in speech .
would you be able to use the ellipses for both purposes in one sentence ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ david ] if you did n't use ellipses , you could just render that however you please , because those were words that were said approximately in that order , right . - [ paige ] yeah , without ellipses or -- - [ david ] without ellipses . - [ paige ] okay .
does there necessarily have to be 3 periods to represent ellipses ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ david ] use number one . it indicates a pause in speech . so like when you said , `` hello , david . ''
so what they are saying at around is that ellipses are the three dot to indicate a pause in speech ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
so , if i quote someone and i just sort of willy-nilly take words out without indicating that i 've taken some information out of the quote , i can make it seem like someone said something totally different to what they actually said , right ? it can be so easy to misconstrue someone 's actual words . so this shows tha...
what is a misconstrue mean ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
those are some of the words he said , but that 's not the idea he was trying to get across . - [ david ] so you have this responsibility with ellipses to do the right thing , to really represent the way that somebody speaks accurately . - [ paige ] yeah , definitely .
not a good man , '' only an informal thing ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark that is actually made up of three periods with spaces in between them , boop , boop , boop . and it has two main uses .
does it matter whether the spaces are there ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ david ] if you did n't use ellipses , you could just render that however you please , because those were words that were said approximately in that order , right . - [ paige ] yeah , without ellipses or -- - [ david ] without ellipses . - [ paige ] okay .
can you explain what ellipses are ?
- [ paige ] hello , grammarians . hello , david . - [ david ] hello . . . paige . - [ paige ] ( chuckles ) so in this video we 're gon na talk about a piece of punctuation called the ellipsis or ellipses in the plural form as we have here . so , what is an ellipsis ? - [ david ] so an ellipsis is a punctuation mark tha...
- [ paige ] hello , grammarians . hello , david .
why is grammar all was changeling ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we do n't know the true parameters p1 and p2 . we do n't know the true population parameters . we do n't know p1 and p2 .
do n't we need to calculate unbiased variance for the two samples in order to estimate the true population variance to be able to calculate the sampling distribution variance ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
that 's part of the problem . we 're trying to figure out if there 's a meaningful difference between p1 and p2 . but we 've seen it multiple times .
as in , should we really just use p1 and p2 to directly calculate the population distribution variance when plugging into the calculation of the difference sampling distribution 's variance ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so this right here is equal to 0.043 . and just like that , we have our confidence interval . we know that there 's a 95 % chance that the true difference of the proportions is within 0.043 of the actual difference of our sample proportions that we got .
if you ended up with partially negative confidence interval would it still be statistically significant if simply a larger portion of it was on the positive side ( thus showing men favoured the `` 1 '' candidate ) ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so let 's do that . so we could have 0.051 minus 0.043 is going to give us 0.008 . and then if we add it , so 0.051 plus 0.043 , it gives us 0.094 . so the 95 % confidence interval between the proportions of men and the proportion of women who are going to vote for the candidate for p1 minus p2 is 0.008 to 0.094 .
from -0.04 to 0.065 at what point does it lose statistical significance ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so if we looked to this normal distribution right over here , this distance that we care about is going to be 1.96 times the standard deviation of this distribution , so it 's going to be 1.96 times all of this business . 1.96 times the standard deviation of this distribution . and so we just need to calculate this and...
why is sal not taking `` corrected standard deviation '' ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval .
the variance presented on the video for the bernoulli distribution is the population variance , however what we have is only a sample , so should n't it be , men for example , s^2_1= ( 642 ( 1-0.642 ) ^2+ ( 1000-642 ) ( 0-0.642 ) ^2 ) /999 ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we do n't know the true parameters p1 and p2 . we do n't know the true population parameters . we do n't know p1 and p2 .
to test if women population is any way different from men population ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so let 's do that . so we could have 0.051 minus 0.043 is going to give us 0.008 . and then if we add it , so 0.051 plus 0.043 , it gives us 0.094 . so the 95 % confidence interval between the proportions of men and the proportion of women who are going to vote for the candidate for p1 minus p2 is 0.008 to 0.094 .
why is difference of 0 to 0.799 % not included ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so let 's just multiply that . 0.022 times 1.96 gives 0.043 . i 'll just round it .
if difference between means is 0 for men and women , does not that mean that they are going to vote for the same person 1 ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so this right here is equal to 0.043 . and just like that , we have our confidence interval . we know that there 's a 95 % chance that the true difference of the proportions is within 0.043 of the actual difference of our sample proportions that we got .
should we not include the values in our confidence interval ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we got 0.642 for the men and 0.591 for the women . but our goal is to get a 95 % confidence interval . so just based on our actual sample , we got -- let me write it over here .
if we are asked to draw a 95 % confidence interval , why can we not just use the empirical rule to know that the mean must be within 2 standard deviations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so that 's this term right over here divided by 1,000 . once again , i need to get the parentheses right . and then we just need to close the parentheses , this original parentheses , because we 're taking the square root of everything .
so to get it right , as a conclusion , the best interest for the the candidate is to increase the adverting which is targeted for women to get more women to vote for him ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
and then if we add it , so 0.051 plus 0.043 , it gives us 0.094 . so the 95 % confidence interval between the proportions of men and the proportion of women who are going to vote for the candidate for p1 minus p2 is 0.008 to 0.094 . i have it right here on the calculator .
how can we conclude that men are more likely to vote for the candidate than women when our confidence interval for difference of means is as low as 0.008 ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
that 's part of the problem . we 're trying to figure out if there 's a meaningful difference between p1 and p2 . but we 've seen it multiple times .
is it correct to conclude that as the confidence intervals of p1-p2 does not containing 0 it imply that there is a statistically significant difference between the populations ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
we do n't know the true parameters p1 and p2 . we do n't know the true population parameters . we do n't know p1 and p2 . that 's part of the problem .
is the application the same if the n 's are different for each sample ( just plugging in the various n 's where applicable ) ?
where we left off in the last video , we were trying to figure out if there 's a meaningful difference between the proportion of men voting for a candidate and the proportion of women . we sampled 1,000 men , sampled 1,000 women , and we got a sample proportion for each of them . we got 0.642 for the men and 0.591 for ...
so this right here is equal to 0.043 . and just like that , we have our confidence interval . we know that there 's a 95 % chance that the true difference of the proportions is within 0.043 of the actual difference of our sample proportions that we got .
why when fining the confidence interval , did we not divide by 10 ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
8 times 2 is 16 is equal to bc times bc -- is equal to bc squared . and so bc is going to be equal to the principal root of 16 , which is 4 . bc is equal to 4 .
is principal root same as the square root of any number ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so these are larger triangles and then this is from the smaller triangle right over here . corresponding sides . and this is a cool problem because bc plays two different roles in both triangles .
is it algebraically possible for a triangle to have negative sides ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and so we know that two triangles that have at least two congruent angles , they 're going to be similar triangles . so we know that triangle abc -- we went from the unlabeled angle , to the yellow right angle , to the orange angle . so let me write it this way .
how do you know that angle b is congruent to angle d ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
this is our orange angle . we know the length of this side right over here is 8 . and we know that the length of this side , which we figured out through this problem is 4 .
i understand how to find a side length , but how do you find the altitude height ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and so bc is going to be equal to the principal root of 16 , which is 4 . bc is equal to 4 . and we 're done .
could n't you just cancel out the bc 's and then find out what ac over dc ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and it 's good because we know what ac , is and we know it dc is . and so we can solve for bc . so i want to take one more step to show you what we just did here , because bc is playing two different roles .
could n't the pythagorean theorem be used to double check whether the answer is correct , or to solve the problem ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so when you look at it , you have a right angle right over here . so in triangle bdc , you have one right angle . in triangle abc , you have another right angle .
can we say a triangle is similar to another triangle if its one side and one angle are same ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
they both share that angle there . and so we know that two triangles that have at least two congruent angles , they 're going to be similar triangles . so we know that triangle abc -- we went from the unlabeled angle , to the yellow right angle , to the orange angle .
how can you tell that the triangles are similar ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so when you look at it , you have a right angle right over here . so in triangle bdc , you have one right angle . in triangle abc , you have another right angle . if we can show that they have another corresponding set of angles are congruent to each other , then we can show that they 're similar .
how are ratios solved without a given number on triangle for any side or angle ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
they both share that angle there . and so we know that two triangles that have at least two congruent angles , they 're going to be similar triangles . so we know that triangle abc -- we went from the unlabeled angle , to the yellow right angle , to the orange angle .
how to prove that is similar triangles ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so in both of these cases . so these are larger triangles and then this is from the smaller triangle right over here . corresponding sides .
although this concept make complete sense with right triangles , would it be applicable to non-right triangles as well ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so you could literally look at the letters . a and c is going to correspond to bc . the first and the third , first and the third .
why does n't a correspond to c and instead b ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
on this first statement right over here , we 're thinking of bc . bc on our smaller triangle corresponds to ac on our larger triangle . and then in the second statement , bc on our larger triangle corresponds to dc on our smaller triangle .
how does bc on the larger triangle correspond with dc on the smaller triangle ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
there 's actually three different triangles that i can see here . this triangle , this triangle , and this larger triangle . if we can establish some similarity here , maybe we can use ratios between sides somehow to figure out what bc is .
. triangle abd does not share that point though ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
let me do that in a different color just to make it different than those right angles . they both share that angle there . and so we know that two triangles that have at least two congruent angles , they 're going to be similar triangles . so we know that triangle abc -- we went from the unlabeled angle , to the yellow...
do you need to have that third angle , c , in order for the three triangles to be similar or are angles b and d enough ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and so we know that two triangles that have at least two congruent angles , they 're going to be similar triangles . so we know that triangle abc -- we went from the unlabeled angle , to the yellow right angle , to the orange angle . so let me write it this way .
at around sal says that angle b is corresponding to angle a , but would it be angle c of the smaller triangle that corresponds with ange a of the biggest triangle ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so if you found this part confusing , i encourage you to try to flip and rotate bdc in such a way that it seems to look a lot like abc . and then this ratio should hopefully make a lot more sense .
how do i solve a right triangle when the altitude , one half of the hypotenuse and one leg are variables ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so when you look at it , you have a right angle right over here . so in triangle bdc , you have one right angle . in triangle abc , you have another right angle . if we can show that they have another corresponding set of angles are congruent to each other , then we can show that they 're similar .
the triangle adb ( the medium one ) is also similar to triangle abc ( the largest ) , right ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and so bc is going to be equal to the principal root of 16 , which is 4 . bc is equal to 4 . and we 're done .
why bc is equal to 4 ( in ) ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so we want to make sure we 're getting the similarity right . white vertex to the 90 degree angle vertex to the orange vertex . that is going to be similar to triangle -- so which is the one that is neither a right angle -- so we 're looking at the smaller triangle right over here .
is vertex abc and vertex bdc are simliar ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
so we want to make sure we 're getting the similarity right . white vertex to the 90 degree angle vertex to the orange vertex . that is going to be similar to triangle -- so which is the one that is neither a right angle -- so we 're looking at the smaller triangle right over here .
what does 'vertex ' mean ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
this is our orange angle . we know the length of this side right over here is 8 . and we know that the length of this side , which we figured out through this problem is 4 .
is there a convenient way to find out which side is congruent with the other side ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
on this first statement right over here , we 're thinking of bc . bc on our smaller triangle corresponds to ac on our larger triangle . and then in the second statement , bc on our larger triangle corresponds to dc on our smaller triangle .
could line segment `` bc `` on the large triangle = line segment `` db `` on the smaller triangle ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
ac is going to be equal to 8 . 6 plus 2 . so we know that ac -- what 's the corresponding side on this triangle right over here ?
: x^2+ ( 1/x^2 ) =6 , what is the value of x- ( 1/x ) ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here .
what do you need to do to achieve the similarity challenge patch ?
in this problem , we 're asked to figure out the length of bc . we have a bunch of triangles here , and some lengths of sides , and a couple of right angles . and so maybe we can establish similarity between some of the triangles . there 's actually three different triangles that i can see here . this triangle , this t...
and so bc is going to be equal to the principal root of 16 , which is 4 . bc is equal to 4 . and we 're done .
( and throughout the whole video ) should n't the measurements of ad and dc be the same ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
all right , so this value is going to be much less than one and that 's how we recognize , that 's one way to recognize a weak acid . look at the ka value .
why is water left out of the ka equation ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
we would form the acetate anions . let me go ahead and draw in the acetate anion so negative one charge on the oxygen . let me show those electrons .
my question is , why is the acetate ion a strong base when the two oxygen atoms and carbon atom can participate in resonance , or electron de-localization to stabilize the negative charge ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so these two electrons in red here are gon na pick up this proton forming this bond . so we make hydronium h30 plus and these electrons in green right here are going to come off onto the a to make a minus . let 's go ahead and draw that in .
should n't resonance make the ion more stable , or am i missing a point here ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
when we write the equilibrium expression , write ka is equal to the concentration of your product so ch3coo minus times the concentration of h3o plus , all over the concentration of acetic acid because we leave water out . so all over the concentration of acetic acid . all right , the equilibrium lies to the left becau...
in the acetic acid and water reaction , can the acetic acid grab a proton from water instead of donating it ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so the stronger the acid , the weaker the conjugate base . water is a much stronger base than the chloride anion . finally let 's look at acetic acids . acetic acid is going to be our bronsted-lowry acid and this is going to be the acidic proton . water is gon na function as a bronsted-lowry base and a lone pair of ele...
because one of the oxygen 's in the acetic acid has two lone pairs and that would be enough to nab a proton from water , no ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
we get approximately 100 % ionization , so everything turns into our products here and let 's go ahead and write our equilibrium expression . so ka is equal to a concentration of h3o plus . so concentration of our products times concentration of cl minus , all over , right , we have hcl and we leave out water . if we t...
did concentration of reactants over the concentration of products ) , would that be your kb ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
hcl is gon na function as a bronsted-lowry acid and donate a proton to water which is going to be our bronsted-lowry base . and so we could think about a loan pair of electrons in the auction taking our proton , leaving those electrons behind . and so the auction is now bonded to three hydrogens because it picked up a ...
when the electrons from water are donated to the hydrogen , is it wrong to think that the hydrogen is attracted to lone pair ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
once ha donates a proton , we 're left with the conjugate base which is a minus . water , h2o accepted a proton , so this is our bronsted-lowry base and then once h2o accepts a proton , we turn into hydronium h3o plus . so this is the conjugate acid .
if h2o is present in a given equation will it always be the blb ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so another way to write this acid base reaction would be just to write acetic acid , ch3 , cooh plus h2o gives us the acetate anion , ch3coo minus plus h3o plus . now acetic acid is a weak acid and weak acids do n't donate protons very well . so acetic acid is gon na stay mostly protonated .
why does n't h40 exist ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so either one is fine . all right and we know when we 're writing an equilibrium expression , we 're gon na put the concentration of products over the concentration of reactants . over here for our products we have h3o plus , so let 's write the concentration of hydronium h3o plus times the concentration of a minus , s...
how do you know if you have more reactants than products ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
all right , so this is a very small number . so if you think about what that does to the ka , all right , a very small number divided by a very large number , this gives you a ka value , an ionization constant much less than one . all right , so this value is going to be much less than one and that 's how we recognize ...
how do i know which one is the smaller number and which is the bigger number in the ka equation ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so we have a very , very large number in the numerator and extremely small number in the denominator . if you think about what that does for your ka , that 's gon na give you an extremely high value for your ka . all right , so ka is much , much , much greater than one here . that 's how we recognize a strong acid .
and also , what happens if ka is less than 1 or greater than 1 ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so let me write that here . the stronger the acid , so stronger the acid , weaker the conjugate , weaker the conjugate base . and one way to think about that is if i look at this reaction , we can think about competing base strength .
weaker the acid , stronger the conjugate base ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
it 's a pure liquid . its concentration does n't change and so we leave , we leave h2o out of our equilibrium expression . all right , so let 's use this idea of writing an ionization constant and let 's apply this to a strong acid .
how is h3o+ formed without change in concentration of h2o ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton .
but would n't it be 2- since oxygen has 6 valence electrones ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
if we think about approximately 100 % ionization , we have all products here . so we have a very , very large number in the numerator and extremely small number in the denominator . if you think about what that does for your ka , that 's gon na give you an extremely high value for your ka .
how do i figure out if the numerator/denuminator is a small or large number ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
when we write the equilibrium expression , write ka is equal to the concentration of your product so ch3coo minus times the concentration of h3o plus , all over the concentration of acetic acid because we leave water out . so all over the concentration of acetic acid . all right , the equilibrium lies to the left becau...
how can you tell that acetic acid is `` not good '' at donating a proton ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
if we think about approximately 100 % ionization , we have all products here . so we have a very , very large number in the numerator and extremely small number in the denominator . if you think about what that does for your ka , that 's gon na give you an extremely high value for your ka .
how do you know that the numerator or denominator is larger just by looking at the molecular formula ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
when you think about this reaction coming to an equilibrium , you 're gon na have a relatively high concentration of your reactants here . when we write the equilibrium expression , write ka is equal to the concentration of your product so ch3coo minus times the concentration of h3o plus , all over the concentration of...
wait , does this mean that ka and pka is related to and works for only if the acid is in water since different liquids other than water can have a different concentration other than 1 in the reaction ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so another way to write this acid base reaction would be just to write acetic acid , ch3 , cooh plus h2o gives us the acetate anion , ch3coo minus plus h3o plus . now acetic acid is a weak acid and weak acids do n't donate protons very well . so acetic acid is gon na stay mostly protonated .
how do we know if it is a strong acid or a weak acid ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
once ha donates a proton , we 're left with the conjugate base which is a minus . water , h2o accepted a proton , so this is our bronsted-lowry base and then once h2o accepts a proton , we turn into hydronium h3o plus . so this is the conjugate acid .
can hydronium accept another proton and turn into oh4+2 ( oxygen tetroxide cation ) ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
and so we write our equilibrium constant and now we 're gon na write ka which we call the acid , the acid ionization constant . so this is the acid ionization constant or you might hear acid dissociation constant , so acid dissociation . so either one is fine .
what is a generic acid ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
base water is acting as a bronsted-lowry base and accepting a proton . and over here if you think about the reverse reaction , the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here but since hcl is so good at donating protons , that means that the chloride anion is not very...
why chlorine anion is not good at accepting protons from hcl ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
strong acids donate protons very easily and so we can say this process occurs 100 % . so we get 100 % ionization . the equilibrium is so far to the right that i just drew this one arrow down over here .
how do i get the numbers ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
finally let 's look at acetic acids . acetic acid is going to be our bronsted-lowry acid and this is going to be the acidic proton . water is gon na function as a bronsted-lowry base and a lone pair of electrons in the auction is going to take this acidic proton , leaving these electrons behind on the oxygen .
how can we tell based on the formula whether an acid is going to be strong or not ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
all right , so h3o plus , so let me go ahead and draw in hydronium . so plus one formal charge on the oxygen and let 's show those electrons in red . all right , so this electron pair picks up the acidic proton .
cl has -1 charge it 's normal charge or formal charge ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so we 're gon na make a minus . let me draw these electrons in green and give this a negative charge like that . let 's analyze what happened .
why did jay place the negative charge in front of oh ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so we 're going to get a very large number for the denominator , for this concentration so this is a very large number and a very small number for the numerator . all right , so this is a very small number . so if you think about what that does to the ka , all right , a very small number divided by a very large number ...
if kb is very very small , so is kw almost zero then ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
and these electrons in green move off onto the chlorine , so let 's show that . we form the chloride anion . let me go ahead and draw in the electrons in green and let me go ahead and write a negative one charge here like that .
why does n't the chloride anion nucleophilicly attack the hydronium cation ?
let 's look at this acid base reaction . so water is gon na function as a base that 's gon na take a proton off of a generic acid ha . so lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the a . oxygen , oxygen is now bonded to three hydrogens . so it picked up a proton . that...
so we 're gon na make a minus . let me draw these electrons in green and give this a negative charge like that . let 's analyze what happened .
why does the cl ion have a negative charge when it has a full octet ?