Answer:
a) Equilibrium shifts in forward direction.
Explanation:
If pressure is increased, equilibrium shifts to the side with the fewer moles of gas.
There are 4 moles of gas in the reactants and 2 moles of gas in the products.
The equilibrium will shift in the forward direction towards the products.
Hope that helps.
<h2>Answer:</h2>
C. Gases only
<h3>Explanation:</h3>
Gases are the state of matter that has no definite shape and volume while liquids have no definite shape but have definite volume and solids have both definite shape and definite volume. so the correct option is "C".
Answer:
0.906 gm/l
Explanation:
We know that molarity of the a solution is given by,
, where 'M' is the molarity, 'w' is the weight of the sample, 'm' is the molar mass, and 'v' is the volume of the solution.
Molarity of a solution tells us the concentration of the solute in the solvent.
Molar mass of KCl is = 74.55
Putting the values we get,
So the molarity of KCl solution is 0.906 gm/l.
The big bang did not produce a significant proportion of elements heavier than helium because the temperatures and densities present in the early universe were not sufficient to support the fusion of heavier elements.
During the first few minute of the big bang, the universe was composed of mostly hydrogen and helium, with very small amounts of lithium and beryllium. As the universe expanded and cooled, the denser regions of the universe collapsed to form the first stars. Inside these stars, the intense pressure and heat generated by nuclear fusion reactions allowed for the production of heavier elements, such as carbon and oxygen. However, elements heavier than helium, such as iron and nickel, require even higher temperatures and densities to be produced, which can only be found in the cores of supernovae. Therefore, the big bang alone did not produce a significant proportion of elements heavier than helium.
to know more about compounds-
brainly.com/question/12166462
#SPJ4
Explanation:
According to Clausius-Claperyon equation,
![ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%20%3D%20%5Cfrac%7B-%5Ctext%7Bheat%20of%20vaporization%7D%7D%7BR%7D%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
The given data is as follows.
= (63.5 + 273) K
= 336.6 K
= (78 + 273) K
= 351 K
= 1 atm, 
= ?
Putting the given values into the above equation as follows.
![ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]](https://tex.z-dn.net/?f=ln%20%28%5Cfrac%7B1.75%20atm%7D%7B1%20atm%7D%29%20%3D%20%5Cfrac%7B-%5Ctext%7Bheat%20of%20vaporization%7D%7D%7B8.314%20J%2Fmol%20K%7D%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7B351%20K%7D%20-%20%5Cfrac%7B1%7D%7B336.6%20K%7D%5D)
= 
= 
= 3813.1 J/mol
Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.
