The ML of 0.85 m NaOH required to titrate 25 ml of 0.72m hbr to the equivalence point is calculated as follows
calculate the moles of HBr used
moles = molarity x volume
25 x0.072/1000= 0.0018 moles
write the equation for reaction
NaOH + HBr = NaBr + H2O
from reacting equation the mole ratio between NaOH to HBr is 1:1 therefore the moles of NaOH = 0.0018 moles
volume = moles/molarity
0.0018/0.085 = 0.021 L in Ml = 0.021 x1000=21.18 Ml ofNaOH
Assuming this is a true or false question, the answer would be True
Answer:
25°C
Explanation:
Combined Gas Law (P₁V₁)/T₁ = (P₂V₂)/T₂
(0.947 atm)(150 mL)/25°C = (0.987 atm)(144mL)/T₂
5.682 = 142.128/T₂
T₂ = 142.128/5.682
T₂ = 25.0137272756°C = 25°C
Answer:
The equilibrium concentrations are:
[SO2]=[NO2] = 0.563 M
[SO3]=[NO] = 1.04 M
Explanation:
<u>Given:</u>
Equilibrium constant K = 3.39
[SO2] = [NO2] = [SO3] = [NO] = 0.800 M
<u>To determine:</u>
The equilibrium concentrations of the above gases
Calculation:
Set-up an ICE table for the given reaction

I 0.800 0.800 0.800 0.800
C -x -x +x +x
E (0.800-x) (0.800-x) (0.800+x) (0.800+x)
The equilibrium constant is given as:
![Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BSO3%5D%5BNO%5D%7D%7B%5BSO2%5D%5BNO2%5D%7D%3D%5Cfrac%7B%280.800%2Bx%29%5E%7B2%7D%7D%7B%280.800-x%29%5E%7B2%7D%7D)

x = 0.2368 M
[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M
[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M