The answer for the following question is mentioned below.
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Explanation:
Given:
Pressure of gas (P) = 1.2 atm
Volume of a gas (V) = 50.0 liters
Temperature (T) =650 K
To calculate:
no of moles present in the gas (n)
We know;
According to the ideal gas equation;
We know;
<u>P × V = n × R × T
</u>
where,
P represents pressure of the gas
V represents volume of the gas
n represents no of the moles of a gas
R represents the universal gas constant
where the value of R is 0.0821 L atm mole^{-1} K^-1
T represents the temperature of the gas
As we have to calculate the no of moles of the gas;
n = 
n = \frac{1.2*50.0}{0.0821*650}
n = \frac{60}{53.365}
n = 1.12 moles
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Answer: I think the answer is C)
Explanation:
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
The PRODUCT is found on the right side of the arrow in a chemical reaction.
The empirical formula CH₂O has a mass [(12 × 1) + (1 × 2) + (16 × 1)] = 30 g/mol
If the empirical formula is 30 g/mol,
and the molecular formula is 60 g/mol
Then the multiple is = 60 g/mol ÷ 30 g/mol
= 2
Therefor the molecular formula is 2(CH₂O) = C₂H₄O₂ (OPTION 2)
