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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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The accepted value for the mass of a moonrock is 46.37 g. In an experiment the rock is measured to be 47.25 g. What is the perce

Artist 52 [7]

Answer:

The answer is

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual measurement = 46.37 g

error = 47.25 - 46.37 = 0.88

The percentage error of the measurement is

$P(\%) = \frac{0.88}{46.37} \times 100 \\ = 1.89777873...$

We have the final answer as

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3 years ago

What volume of carbon dioxide gas can be collected from

alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we determine which reactant is limiting:

Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃

Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

0.05 mol HCl * $\frac{1molCO_{2}}{1molHCl}$ = 0.05 mol CO₂

Finally we use PV=nRT to calculate the volume:

P = 1 atm

n = 0.05 mol

T = 25 °C ⇒ 25 + 273.16 = 298.16 K

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3 years ago

Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)

olganol [36]

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calculation

2 Al + Fe2O3 → 2Fe + Al2O3

step 1: find the moles of Al by use of moles= mass/molar mass formula

= 60.0/27= 2.22 moles

Step 2: use the mole ratio to determine the moles of Al2O3.

The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles

Step 3: finds the mass of Al2O3 by us of mass= moles x molar mass formula.

The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol

mass is therefore= 102 g/mol x 1.11= 113.22 grams

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3 years ago

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ra1l [238]

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