Answer:
Concentration of product at equilibrium ;
![[H^+]=0.0000229 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.0000229%20M)
![[CN^-]=0.0000229 M](https://tex.z-dn.net/?f=%5BCN%5E-%5D%3D0.0000229%20M)
Explanation:
initially
0.85 M 0 0
(0.85-x)M x x
The equilibrium constant of reaction = 
The expression of an equilibrium cannot can be written as:
![K_c=\frac{[H^+][CN^-]}{[HCN]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D)
Solving for x:
x = 0.0000229
Concentration of product at equilibrium ;
Your answer would be,
Molarity = moles of solute/volume of solution we needed, 29.22(g)(mol) of NaCI
= 29.22(g)/58.44(g)(mol^-1)(1)/1(L)
= 0.500(mol)(L^-1)
Hope that helps!!!
Answer:
A physical property that depends on the sample size
Explanation:
Answer: 2.17 x 10^23 molecules
Explanation:
1mole of H2O contains 6.02x10^23 molecules.
Therefore 0.360mole of H2O will contain = 0.36 x 6.02x10^23 = 2.17 x 10^23 molecules
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
