Question

1. how many hydrogen atoms are present in one formula unit of ammonium hydrogen phosphate, $(nh_4)_2hpo_4$?a. 4b. 8c. 9d. 167. the smallest ratio of elements in the compound sodium peroxide, $na_2o_2$, isa. 1:1.b. 2:1.c. 1:2.d. 2:2.8. the molar mass of manganese, mn, isa. 25 amu.b. 25 g/mol.c. 54.94 amu.d. 54.94 g/mol.9. the molar mass of sodium fluoride, naf, isa. 20.00 g/mol.b. 41.99 g/mol.c. 60.99 g/mol.d. 64.98 g/mol.10. the molar mass of $al_2(so_4)_3$ isa. 75.05 g/mol.b. 150.03 g/mol.c. 262.17 g/mol.d. 342.17 g/mol.

Explanation:

Question 6 Step1: Count H in $\text{NH}_4$

Each $\text{NH}_4$ has 4 H atoms.

Question 6 Step2: Multiply by subscript 2

$4 \times 2 = 8$
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Question 7 Step1: Simplify $\text{Na}_2\text{O}_2$ ratio

Divide 2:2 by 2, get 1:1.
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Question 8 Step1: Recall Mn molar mass

Molar mass of Mn is 54.94 g/mol.
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Question 9 Step1: Find atomic masses

Na: 22.99 g/mol, F: 19.00 g/mol.

Question 9 Step2: Sum for NaF

$22.99 + 19.00 = 41.99$ g/mol
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Question 10 Step1: List atomic masses

Al: 26.98 g/mol, S: 32.07 g/mol, O: 16.00 g/mol.

Question 10 Step2: Calculate total mass

$$(2 \times 26.98) + (3 \times 32.07) + (12 \times 16.00)$$
$$= 53.96 + 96.21 + 192.00 = 342.17$$ g/mol

Answer:

1. b. 8
2. a. 1:1.
3. d. 54.94 g/mol.
4. b. 41.99 g/mol.
5. d. 342.17 g/mol.