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3 years ago

What can make this balanced by adding coefficients CH4+O2 H2O+ CO2

Chemistry

2 answers:

Ugo [173]3 years ago

6 0

CH4+2O2=> 2H2O+CO2

C- 1 on both sides
H- 4 on both sides
O - 4 on both sides

user100 [1]3 years ago

3 0

Answer:

$\huge \boxed{\mathrm{CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O}}$

$\rule[225]{225}{2}$

Explanation:

$\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O$

Balancing the Hydrogen atoms on the right side,

$\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O$

Balancing the Oxygen atoms on the left side,

$\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O$

$\rule[225]{225}{2}$

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3 years ago

A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

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Answer:

The empirical formula of the compound = $C_4H_8S_1O_1$

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide = $\frac{0.059060 g}{44 g/mol}=0.0013 mol$

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Moles of water = $\frac{0.024176 g}{18 g/mol}=0.0013 mol$

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen= $1 g/mol\times 0.0013 mol=0.0013 g$

Mass of sulfur dioxide = 20.326 mg = 0.020326 g

Moles of sulfur dioxide = $\frac{0.020326 g}{64 g/mol}=0.00032 mol$

Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol

Mass of 0.00032 moles of sulfur = $32 g/mol\times 0.00032 mol=0.01024 g$

Mass of oxygen in the sample = x

Mass of sample = 33.153 mg = 0.033153 g

0.033153 g = 0.0156 g + 0.0013 g + 0.01024 g + x

x = 0.006013 g

Moles of oxygen = $\frac{0.006013 g}{16 g/mol}=0.00038 mol$

For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:

Carbon : $\frac{0.0013 mol}{0.00032 mol}=4$

Hydrogen: $\frac{0.0026 mol}{0.00032 mol}=8$

Sulfur : $\frac{0.00032 mol}{0.00032 mol}=1$

Oxygen : $\frac{0.00038 mol}{0.00032 mol}=1$

The empirical formula of the compound = $C_4H_8S_1O_1$

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4 years ago

If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at a constant temperature. What is the new volume

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Answer :

=748

⋅22.5

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