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Problem 4

University of Toronto

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Problem 1

The atomic mass of $\mathrm{Cl}$ is 35.45 amu, and the atomic mass of Al is 26.98 amu. What are the masses in grams of 3 $\mathrm{mol}$ of $\mathrm{Al}$ atoms and of 2 $\mathrm{mol}$ of $\mathrm{Cl}$ atoms?

Answer

$\operatorname{Ar}(\mathrm{Al})=26,98$ amu

$\mathrm{n}(\mathrm{Al})=3 \mathrm{mol}$

$\mathrm{m}=\mathrm{n}^{*} \mathrm{Ar}(\mathrm{Al})=26.98 \mathrm{g} / \mathrm{mol} * 3 \mathrm{mol}=80,94 \mathrm{g}$

$\operatorname{Ar}(\mathrm{Cl})=35,45 \mathrm{amu}$

$\mathrm{n}(\mathrm{Cl})=2 \mathrm{mol}$

$\mathrm{m}=\mathrm{n}^{*} \mathrm{Ar}(\mathrm{Cl})=35,45 \mathrm{g} / \mathrm{mol} * 2 \mathrm{mol}=70,9 \mathrm{g}$

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## Discussion

## Video Transcript

for they Thomas or atomic mass of Korean Element. They accost, pointing to three or 5.45 AM Jew and 26.9 seats AM do you? So what would be the Massport to move off? Cory Natural are so free most over me. Um so, first of all, we have to know are we to convert atomic mass to morning mass? So for one, am you essentially it's actually used equals to one gram per artistic conversion factors, so we can definitely use Ah, um the at a massive Adebola mass. And then we came out of veteran by number almost to find out the mass for two mo off Kar Kar Corey Sofa Corinne And we have 35.45 um, gram per mo We convert am due to, ah, gram per mo chance to. So from here, we should be able to find it. Um, it would be because of 70 point now I graham, because we have to mow. So two more over here. So the unique as l you have 79 70.9 grams for l Let me, um we go just just free most. My reply by them will amass 26.8 gram per mo Ah, you will be closed to 26.8 times 3 89 80.94 Grab and here address.

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