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1- Complete the following proof. Theorem Let S ER: Let p € R The following two statements are equitlent: (i) Ve > Ve(p)o (S| {p}) "onemptu: (ii) P= l...

Question

1- Complete the following proof. Theorem Let S ER: Let p € R The following two statements are equitlent: (i) Ve > Ve(p)o (S| {p}) "onemptu: (ii) P= lim dn for some sequence (an) contained in $ | {p} (Recall that if any of the above equivalent conditions holds, then we say that is a limit poinf of $.) Proof: (ii) Let € be given_ Qur goal , is to show that Ve(p)n(S| {p}) is nonempty: We have Uj `p JN €N Vn >N dn € Velp) In particular; aN+I Ve(p) . Also notice that; by assumpti

1- Complete the following proof. Theorem Let S ER: Let p € R The following two statements are equitlent: (i) Ve > Ve(p)o (S| {p}) "onemptu: (ii) P= lim dn for some sequence (an) contained in $ | {p} (Recall that if any of the above equivalent conditions holds, then we say that is a limit poinf of $.) Proof: (ii) Let € be given_ Qur goal , is to show that Ve(p)n(S| {p}) is nonempty: We have Uj `p JN €N Vn >N dn € Velp) In particular; aN+I Ve(p) . Also notice that; by assumption aN+1 €s {p}: This shows that Ve(p)n(s| {p}) is nonempty: (complete the proof. Hint: Notice that for all €N; by assumption; (p- #p+4) has nonempty intersection with {p} Construct sequence (dn) by letting On be point in the nonempty intersection of (p 2P+ and {p}- Use the Squeeze Theorem to show that indeed &r - 2- Let sequence of real numbers. Let s = {an : " €N} Prove that if L is limit point of (that is, if L € $'), then there exists subsequence of (an) whose limit is L Proof: In the reading assignment posted to the course homework page we proved the following lemma: Lemma. Let E CR Let p € E' and suppose that 0 is aJ open set containing Tlen On_ contains infinitely aiI pomts Since L € $', it follows from the above lemma that (L-LLtI)nshas infinitely many points 3n such that dnp (L-LL+I) L+ ins has infinitely many points 302` "g such that arz € (L = ~L+ Lt nshas infinitely many points 3n3 nz such that Ors 3-L+ In ths way we obtain subsequence Un such that VkeN Sam < Lt} (complete the proof. Hint: Use the Squeeze Theorem to show that lim Ont = L)



Answers

1. Convert the following masses into number of moles
a. 128.96 g of Cu
b. 87.2 g of Hg.
2. The molecular formula of acetylsalicylic acid (aspirin), one of the most common pain
relievers, is HC9H7O4.
a. How many moles of HC9H7O4 are in a 0.500-g tablet of aspirin?
b. How many molecules of Hc9H7O4 are in this tablet?
3. Calculate the percentage by mass of the indicated element in the following compounds:
a. Carbon in acetylene, C2H2, a gas used in welding.
b. Oxygen in ascorbic acid, HC6H7O6, also known as vitamin C
4. The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction
between sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7):
3NaHCO3(aq) + H3C6H5O7 (aq) ? 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric are allowed to react.
a. Which is the limiting reactant?
b. How many grams of carbon dioxide form?
c. How many grams of the excess reactant remain after the limiting reactant is
completely consumed?
5. What is the molarity of potassium hydroxide solution if 38.65 ml of the KOH solution is
required to titrate 25.84 ml of 0.1982 M of hydrochloric acid solution?
6. In an experiment, a student titrated 0.1060 g of sodium carbonate (Na2CO3) dissolved in
water with a standard 0.09115 mol dm-3 hydrochloric acid (HCl) solution from a burette.
a. Write a balanced chemical equation for the titration reaction.
b. What volume of the HCl solution is required to reach the end point?
7. 35 mL of 18.0 M H2SO4 solution was dissolved in enough water to make 250 mL of solution.
What is the molarity of the dilute solution?

This problem covers limits. In order to solve part A and B of this problem, we will use the approach of contradiction. Let's look at part A. First. The claim is that if S. N. Is greater than equal to A. For all but finite lee many N than limit, S N is greater than equal to A. Let us assume for a contradiction that this limit Yeah, is less than let us assume. And let's say absalon is equal to a minus S. This is the assumption. By the definition of limits, we can say that. Bye. Definition of limit. We can see that there exists capital N. Substantive small and is greater than capital in then sn minus S is less than epsilon. This is what the definition says. I'll write it down there exists capital and such that Yeah. If and is greater than capital N. Then and send minus S. Yes, less than absolute. So if you open this inequality, you know, you will get minus epsilon Nice between s and minus S taking minus s on both sides. You get s minus absalon as less absalon. Now, you can see here that S plus epsilon is nothing but A. So you will use that factor and I can say that s minus epsilon. Uh it means that essen belongs to s minus absalon. Call my A. What does this mean? It means that S. N. Is less than a, S. N is less than A. But our problems said that our assumption was based that S and will be greater than equal to way, but we got the opposite. It means that for all but finite lee many end S. N. Is less than in this problem, as per the assumption, which is a contradiction. It means that our initial claim was correct that if S. N. Is greater than equal to A. For all but financially many in limit S and will be greater than equal to It similarly will solve part B of the problem. Part B says that if S. N is less than equal to be for all, but finitely many end then limit essen is less than equal to be. Let us assume again for contradiction. We will assume, Oh, that limit essen is greater than B. We're assuming as a contradiction. And let us take absalon as S minus B. Again, by the definition of limit there exists and such that if N is greater than in the sm minus S is less than absolute, just like this. We use their definition of LTD so we get yes and minus S is less than epsilon. Again, opening the inequality minus epsilon, S n minus S rather than absolute, this becomes s minus epsilon. Essen? S less. Absolutely. From here, we can see that S many cepsa loans nothing but B. So I can see here, B essen less than s plus absalon. So from here it is clearly seen that sns greater than B. Or I can say for all. But finitely many N. We have S. And greater than B, which is a contradiction according to our statement. Because we took us and less than equal to be. Hence the claim that if S. N. Is greater than less than equal to be for all. But finitely many end then limit us and less than equal to be was correct. And contradiction proved it false. Let's go to part C. Part C is a direct conclusion from part A and part B from part A. You know that? Yes. Which is the limit limit of essen is greater than equal to A. And from part B. This limit S it's less than equal to me, so S belongs to it to be. That's all. Yeah.

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

In this problem were given that a sequence essen is such that The magnitude of difference between the end plus one term and the entire term is always less than do you raise to the power of negative and so we can consider the maximum difference rather the magnitude of the maximum difference. Mhm. Being equal to s to the Brother, 2 to the power negative end. Now a cautious sequence is defined as a sequence in which the subject subsequent terms become arbitrarily close to each other. So If a seven is a sequence than the magnitude of a plus and And a seven plus 1 minus a seven, mhm is equal to zero as N approaches infinity. So they become arbitrarily close to each other. Now to prove that this sequence is a cautious sequence. We can dig the limit as N tends to infinity on both sides. Yeah. Mhm Okay, okay, Okay. Mhm. What people saw limit uh and tends to infinity of due to the negative end and we can solve for the limit. What the fuck? Mhm. Okay. And it's actually a pretty easy limit to solve. So that's the limit as intends to infinity of two to the end. Now, as N approaches infinity, due to the end also approaches infinity, so and is in the in the exponent and as and increases due to the end blows up to infinity and that means that one over to the end. Yeah, Ghost zero. So our limit evaluates 20. Yeah. So the limit as n tends to infinity As 7-plus 1 minus S seven Rather their magnitude equal zero and therefore as intends to infinity the subsequent terms right, subsequent terms Of the sequence, S seven become arbitrarily close, become are bit rarely close, and therefore the given sequence is a cause she sequence. And since cautious sequences always can converge so cause she sequences in words follows that the sequence S N. Don't forget that also. Yeah. Okay. And bridges and there is our solution to this broad.

So in this question we have to find out the number of moles for two things. One is see you and another risk X. G. It's in the first case we are seeing. So the given masks that we have is It is given to be won 28.96 grand. And the molar mass of see you. It is equal to 63 point 54. Okay, pottering drugs. So in order to find out the number of moles we can divide these things right? We're going to divide. Mhm. The given mass upon the mamas and that is 1 28.96 divided by 63 points five or six. And that would be coming to about two point 0- 9. 3. So these many moles are there in c the second case that we have that is for HCG. So here the human mosque equal stupid 87.2 g. And the mamas equal soup 205. 205 09 Grand. Therefore we get The number of moles to B equals two. That is given by given mass upon Malema's. That is 87.2 Divided by 200 points 59. And that will be giving us 87.2 divided by 200.59. And that is equal to 0.43 47. Voice. So this is what how we calculate the value. I hope you


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