Answer - 2

Hi! Could you please include the answers from the textbook?

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9 . n=14,6/146=0,1 mol

n of oxygen = (16 * 3 * 2)/16 = 6 mol ( i don’t understand why n is found with this formula)

n° of atoms of oxygen= 0,1 * 6 * Na = 0,6 Na

8 . 1 g molecule = 1 mol so:

1 mol of ideal gas = 22,4L

1 mol = 6 * 10²³ molecules = Na

N2 = 14 * 2 = 28g/mol

3 . gaseous oxygen and nitrogen = O2 and N2

if we have 1 kg of mixture we’ll have 1/5 = weight of O2 = 200g

and 4/5 = weight of N2 = 800g

ratio O2 : N2

200g : 800g

100g : 400g

n=100/32 : n=100/7

n° molecules=7 * 100 * Na : n° molecules=32 * 100 * Na

7 : 32 is the simplified ratio

10 .

1- n=4,25/17=0,25

n° atoms = 0,25 * 4 * Na = 1 Na = 6 * 10²³

2- n=4,2/14=0,3

n° atoms = 0,3 Na

each atom has 8 valence electrons so 8 * 0,3 Na = 2,4 Na

3- molar volume at STP=22,4L

n=V/22,4 = 89,6/22,4 = 4

n° molecules = 4 * Na

I hope this helps, let me know if you have any questions, even though this isn’t my strongest subject

39 . Keep in mind the question is asking for the number of atoms and not molecules!

1 - n=m/M = 2,5/28 = 0,1

n° of atoms = 0,1 * 2 * Na= 1,2 * 10²³

2 - 5/14 = 0,3

n° of atoms = 0,3 * Na = 1,8 * 10²³

3 - 3,0 * 10²⁴ = 30 * 10²³

4 - 82/28 = 3

n° atoms = 3 * 2 *Na = 36 * 10²³ which is the max number of atoms.