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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The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
Answer:
the reaction will shift towards the “heat”—shifts to the left
Explanation:
To summarize:
o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the
endothermic direction.
o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the
exothermic direction.
o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is
towards the “heat” term.
Therefore the reaction will shift towards the “heat”—shifts to the left
Answer:
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