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Klio2033 [76]
2 years ago
6

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

Chemistry
2 answers:
Angelina_Jolie [31]2 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]2 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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Explanation:

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Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)

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If the reactants are 10 kg what is the mass of the products
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7 0
2 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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