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3 years ago

What is the percent by mass of carbon in acetone, c 3 h 6 o?

Chemistry

2 answers:

Angelina_Jolie [31]3 years ago

6 0

Answer: I just took the test and got 100%

1. B: 0.430 mol

2. B: percent composition

3. C: 71.5%

4. B: 62.1%

5. A: empirical formula

6. D: C16H24O4

7. D: 673g

Feel free to mark as brainliest :)

Tatiana [17]3 years ago

5 0

In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C= $( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%$
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%

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Answer:

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Mass of lead acetate = 0 grams

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 1.81 grams

Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.90 grams

Molar mass of sulfuric acid = 98.08 g/mol

Mass of lead acetate = 4.90 grams

Molar mass of lead acetate = 325.29 g/mol

Step 2: The balanced equation

H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH

Step 3: Calculate moles

Moles = mass / molar mass

Moles H2SO4 = 4.90 grams / 98.08 g/mol

Moles H2SO4 = 0.0500 moles

Moles lead acetate = 4.9 grams / 325.29 g/mol

Moles lead acetate = 0.0151 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles

Step 5: Calculate moles of products

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH

Step 6: Calculate mass

Mass = moles * molar mass

Mass H2SO4 = 0.0349 moles * 98.08 g/mol

Mass H2SO4 = 3.42 grams

Mass PbSO4 = 0.0151 moles * 303.26 g/mol

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 0.0302 moles * 60.05 g/mol

Mass of CH3COOH = 1.81 grams

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Answer:

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Given parameters:

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To solve this problem, we use the ideal gas equation:

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Answer:

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Saturated fat -

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