The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.
8. 63360
9. 1000
if I got anything wrong please let me know thank you :)
Answer:

Explanation:
Hello,
In this case, the reaction is:

Thus, the law of mass action turns out:
![Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_3CH_2OH%5D_%7Beq%7D%7D%7B%5BH_2O%5D_%7Beq%7D%5BCH_2CH_2%5D_%7Beq%7D%7D)
Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change
result:
![[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol](https://tex.z-dn.net/?f=%5BCH_2CH_2%5D_%7Beq%7D%3D29mol-x%3D16mol%5C%5Cx%3D29-16%3D13mol)
In such a way, the equilibrium constant is then:

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

Thus, the second change,
finally result (solving by solver or quadratic equation):

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

Best regards.
<u>Given:</u>
Mass of the ball = 35.2 kg
Momentum = 218 kg m/s
<u>To determine:</u>
The velocity of the ball
<u>Explanation:</u>
Momentum (p) of an object is the product of its mass (m) and velocity (v)
p = m*v
v = p/m = 218 kg.ms-1/35.2 kg = 6.19 m/s
Ans: The velocity of the ball is 6.19 m/s