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University of Toronto

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Problem 13

Calculate each of the following quantities:

(a) Amount (mol) of Mn atoms in 62.0 $\mathrm{mg}$ of Mn

(b) Amount (mol) for $1.36 \times 10^{22}$ atoms of $\mathrm{Cu}$

(c) Mass ( g) of $8.05 \times 10^{24}$ Li atoms

Answer

a. $\mathrm{n}(\mathrm{Mn})=0,001 \mathrm{mol}$

b. $\mathrm{n}(\mathrm{Cu})=0,023 \mathrm{mol}$

c. $\mathrm{m}(\mathrm{Li})=113,49 \mathrm{g}$

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## Discussion

## Video Transcript

So we have to find out the lumber more. Magnus in 62 Mini Grandma Maggie's. Okay, Fistful When we ah have to convert mass tourism promo, we're using the equation. It's simple. Most equal to mass over more than mass. So and men stand firm with a mask. And for in this equation for the massive he always in Quem when the master will be graham from bowl and four number more, of course, will be number most. But we're giving these in many Graham, Minnie quay. So for each minute, Graham. So what? Me, Graham. Your course Morning to turn to lethally graham. Oh, um one over 1000 gram. So we had the first convert our weight of the mass from Billy Graham to Graham. So from here we will have ah, the world by 1000 soup on 062 grams, We be our, um mass foam agonies. Okay. And then we're going to found Ah Magan is ah with a mass. So you'll be fitted for point is six. Um sorry. 64.94 Graham Permal And then we we Yeah, maybe just a calculator. We are going to find out. Ah, number most so it will be because the 1.1 time 10 to the letter flee most and his number most off Magallanes in 62 milligrams of Magnus. OK, Poppy, What would be a little more off 1.36 times in the power 22 ATS him or copper? So if the comment from fatter bottom of malware or an atom instead, we'll have one more. We will have. Ah, 64 No, two time. Sandra 2040. That's him. So there's called Abby Gradual number. Yeah. So you want to find the number one was free, you see? So essentially, the number most would be equals our a lumber at him. Divide the biter. I regard your number. So it's 123. And then again, we're going to do so. I'll call later, um to find out. Ah, that your love bo most. So you will be equals two 0.2 to 6. Small for copper. Okay, for ah policy, we have them. We need to find a massive extra point old five times 10 to 24 flew leaving atom. So we're going to look at the page over years, so we need eight point old five times. 10 to 24. Um, Levi, um, that's him. And we want about a mass of it. All right again. You've wanted fire the mast. We're going to look at the number one mass, and we should be able to find that the massacres to limbo most times to remove the mass. All right, so we just need your father number almost so limbo moves again. You ever do close to our atom? Divide the biter. I regard your number six for No. Two. I turned to the power of 24 and then again, you sell Collate over. You should be able to find out. Um uh, number most Fanny should be of off the 13.4. No. Okay, so it's again to adopt double check. Make sure we have there. Right? Answers. Okay, So when I have the lumber almost reaches them. Applied that by the more the mass. So were the mass. ICO still number. Move 30 point for most when applied by the morning myself leaving. So for leafy, um, the moment mass would be 6.9 for one gram for more. You can find out from your task book This number of very little bit. Even what Countess spoke to use them, what would be the source? So, yeah, this is a very over my task book, so we just take 6.9 for one time Virgin point for Ah, we will have 98th week 0.0, gram for 8.5.

## Recommended Questions

Calculate each of the following:

a. number of $\mathrm{Li}$ atoms in 4.5 moles of $\mathrm{Li}$

b. number of $\mathrm{CO}_{2}$ molecules in 0.0180 mole of $\mathrm{CO}_{2}$

c. moles of $\mathrm{Cu}$ in $7.8 \times 10^{21}$ atoms of $\mathrm{Cu}$

d. moles of $\mathrm{C}_{2} \mathrm{H}_{6}$ in $3.75 \times 10^{23}$ molecules of $\mathrm{C}_{2} \mathrm{H}_{6}$

What is the mass in grams of each of the following?

a. $6.02 \times 10^{24}$ atoms $\mathrm{Bi}$

b. $1.00 \times 10^{24}$ atoms $\mathrm{Mn}$

c. $3.40 \times 10^{22}$ atoms He

d. $1.50 \times 10^{15}$ atoms $\mathrm{N}$

e. $1.50 \times 10^{15}$ atoms $\mathrm{U}$

Determine the number of atoms in each of the following.

(Chapter 10)

\begin{equation}

\text { a. }56.1 \mathrm{g} \text { Al } \quad \text { b. } 2 \text { moles } \mathrm{C}

\end{equation}

Calculate each of the following quantities:

(a) Mass (g) of $6.44 \times 10^{-2} \mathrm{mol}$ of $\mathrm{MnSO}_{4}$

(b) Amount (mol) of compound in 15.8 $\mathrm{kg}$ of $\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}$

(c) Number of $\mathrm{N}$ atoms in 92.6 $\mathrm{mg}$ of $\mathrm{NH}_{4} \mathrm{NO}_{2}$

Calculate each of the following quantities:

(a) Mass ( $\mathrm{g}$ ) of 0.68 $\mathrm{mol}$ of $\mathrm{KMnO}_{4}$

(b) Amount (mol) of $\mathrm{O}$ atoms in 8.18 $\mathrm{g}$ of $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$

(c) Number of $\mathrm{O}$ atoms in $7.3 \times 10^{-3} \mathrm{g}$ of $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}$

What is the mass in grams of each of the following? (Hint: See Sample Problems B and E.)

a. 1.00 mol $\mathrm{Li}$

b. 1.00 mol $\mathrm{Al}$

c. 1.00 molar mass $\mathrm{Ca}$

d. 1.00 molar mass $\mathrm{Fe}$

e. $6.022 \times 10^{23}$ atoms $\mathrm{C}$

f. $6.022 \times 10^{23}$ atoms $\mathrm{Ag}$

How many moles contain each of the following?

a. $5.75 \times 10^{24}$ atoms Al $\quad$ b. $2.50 \times 10^{20}$ atoms Fe

Calculate each of the following quantities:

(a) Mass (g) of 0.346 mol of Zn

(b) Number of $\mathrm{F}$ atoms in 2.62 $\mathrm{mol}$ of $\mathrm{F}_{2}$

(c) Number of $\mathrm{Ca}$ atoms in 28.5 $\mathrm{g}$ of $\mathrm{Ca}$

Calculate the mass, in grams, for each of the following:

a. 2.00 moles of $\mathrm{Na}$

b. 2.80 moles of $\mathrm{Ca}$

c. 0.125 mole of $\mathrm{Sn}$

d. 1.76 moles of $\mathrm{Cu}$

What is the mass in grams of each of the following?

a. $3.011 \times 10^{23}$ atoms $\mathrm{F}$

b. $1.50 \times 10^{23}$ atoms $\mathrm{Mg}$

c. $4.50 \times 10^{12}$ atoms $\mathrm{Cl}$

d. $8.42 \times 10^{18}$ atoms $\mathrm{Br}$

e. 25 atoms $\mathrm{W}$

f. 1 atom $\mathrm{Au}$