Answer:
a. Oxygen gas is limiting
Explanation:
hydrogen gas and oxygen gas are reacted to form water
2H₂ + O₂ → 2H₂O
the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂
Given equal masses of H₂ and O₂
assuming 'x' gm for each, no. of moles of each gas =
no. of moles of H₂ = x/2 = 0.5x moles
no.of moles of O₂ = x/32 = 0.031x moles
This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.
4. NaOH
____________________________________
Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>

3.8 /

=

((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
Answer : The correct expression for equilibrium constant will be:
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,

The expression of
will be,
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Therefore, the correct expression for equilibrium constant will be, ![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)