V(HNO₃) = 50.0 mL in liters = 50.0 /1000 =0.05 L
M(HNO₃) = 1.50 M
Number of moles HNO₃ :
n = M x V
n = 1.50 x 0.05
n = 0.075 moles of HNO₃
HNO₃ + NaOH = H₂O + NaNO₃
1 mole HNO₃ -------- ---1 mole NaOH
0.075 moles HNO₃ ---- ?
moles NaOH = 0.075 * 1 / 1
= 0.075 moles of NaOH
V ( NaOH ) :
M = n / V
0.81 = 0.075 / V
V = 0.075 / 0.81
V =<span> 0.0925 L or 92.5 mL </span>
<span>hope this helps!</span>
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:
The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:
Expression for is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Answer:
2
Explanation:
The subscript on Ammoniumwhich is (NH4) is 2.
Hope this helped!
The balanced chemical equation for the reaction is as below
KBr (aq) + AgNo3 → KNO3(aq) + AgBr(s)
Explanation
from the equation above the stoichiometric coefficient before KBr is 1, before AgNo3 is 1 , before KNO3 is 1 and before AgBr is also 1. Therefore 1 mole of KBr reacted with 1 mole of AgNO3 to form 1 mole of KNO3 and 1 mole of AgBr.
The correction according to the Van Dee Walls equation is Vn-b
