The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:
true
Explanation:
This is because it helps distribute the particles that are being dissolved.
3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)
This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.
Hope this helped!
The alloy has a density of 21.186g/cc. So for a kilogram or 1000 grams/21.186 g/cc= 45.7 cc. So the answer is 45.7 cc of the allow to make up a kilogram which shows that the density of the allow can be used to calculate the volume of a larger mass ie the kilogram.
2NaClO3 = 2NaCl + 3O2
Hopes this helps <33
