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:
The second answer, because when something saturated, it has the maximum possible number of hydrogen atoms.
Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
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Answer is: reaction is second order with respect to a.
This second order reaction<span> is proportional to the square of the concentration of reactant a.
</span>rate of reaction = k[a]².
k is second order rate constant and have unit M⁻¹·s⁻¹.
Integrated rate law for this reaction: <span><span>1/[a]</span>=<span>1/<span>[a]</span></span></span>₀ <span>+ kt.
t is time in seconds..</span>
