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Hunter-Best [27]
2 years ago
8

(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.

Chemistry
1 answer:
svp [43]2 years ago
5 0

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

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What is the empirical formula for a compound which contains 67.1 zinc and the rest is oxygen
wolverine [178]

Answer:

The empirical formula is ZnO2

Explanation:

What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Step 1: Data given

Suppose the compound has a mass of 100.0 grams

A compound contains:

67.1 % Zinc  = 67.1 grams

100 - 67.1 = 32.9 % oxygen  = 32.9 grams

Molar mass of Zinc = 65.38 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles of Zinc

Suppose the compound is 100 grams

Moles Zn = 67. 10 grams / 65.38 g/mol

Moles Zn = 1.026 moles

Step 3: Calculate moles of O

Moles O = 32.90 grams / 16.00 g/mol

Moles O = 2.056 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Zn: 1.026/1.026 = 1

O: 2.056/1.026 = 2

The empirical formula is ZnO2

To control this we can calculate the % Zinc for 1 mol

65.38 / (65.38+2*16) = 0.67.1 = 67.2 %

7 0
3 years ago
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