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Ivenika [448]
3 years ago
11

0.3287 g x 45.2 g =?

Chemistry
2 answers:
Mariana [72]3 years ago
8 0

Answer:

14.85724 grams.

Explanation:

murzikaleks [220]3 years ago
4 0

Answer:

The exact answer is 14.85724 grams. Just simply multiply these two numbers together to get your answer in grams. No fancy formula is needed for this! Hopefully this helps!

Feel free to mark brainliest!

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What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?
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Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

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Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

PV=nRT.

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}

The volume would be 158 L.

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