The answer is CH3CH2CH2CHO
Using ideal gas equation,

Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=1 atm
T=25 C+273 K =298.15K
V=663 ml=0.663L
R=0.0821 atm L mol ⁻¹
Mass of gas given=1.25 g g
Molar mass of gas given=?


Putting all the values in the above equation,

Molar mass of the gas=46.15
Answer:

Explanation:
25. Boyle's Law
The temperature and amount of gas are constant, so we can use Boyle’s Law.

Data:

Calculations:

26. Ideal Gas Law
We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.
pV = nRT
Data:
p = 101.3 kPa
V = 20 L
n = 5 mol
R = 8.314 kPa·L·K⁻¹mol⁻¹
Calculation:
101.3 × 20 = 5 × 8.314 × T
2026 = 41.57T

<span>Catalysts decrease the activation energy and the more collisions result in a </span>reaction<span>, so the </span>rate<span> of </span><span>reaction increases.</span><span />
<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.
<u>Explanation:</u>
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

Where,
= change in Gibb's free energy
= change in enthalpy
T = temperature
= change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and
is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the
is also negative.
For a reaction to be spontaneous,
must be negative.
![-ve=-ve-[T(-ve)]\\\\-ve=-ve+T](https://tex.z-dn.net/?f=-ve%3D-ve-%5BT%28-ve%29%5D%5C%5C%5C%5C-ve%3D-ve%2BT)
From above equations, it is visible that
will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.