Answer:
B.0.2 J/g°C
Explanation:
From the attached picture;
- Heat attained in the solid phase is 200 Joules
- Change in temperature is 50°C ( from 0°C to 50°C)
- Mass of the solid is 20 g
We are required to determine the specific heat capacity of the substance;
- We need to know that Quantity of heat is given by the product of mass,specific heat capacity and change in temperature.
- That is; Q = mcΔT
Rearranging the formula;
c = Q ÷ mΔT
Therefore;
Specific heat = 200 J ÷ (20 g × 50°c)
= 0.2 J/g°C
Thus, the specific heat of the solid is 0.2 J/g°C
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.
Answer:
The position of equilibrium would not be appreciably affected by changes in the volume of the container for NiO(s) + CO(g) ⇌ Ni(s) + CO2(g).
Correct Answer : Option A
Explanation:
The equilibrium position tends to change with increase or decrease in pressure or volume, or both of them. This happens because considering change in volume, when the volume of the container increases, the reactant molecule increases i.e. mole of gases and thus the position of equilibrium shifts towards the right side. Similarly in case of decreasing volume, reactant molecule decreases and the equilibrium position shifts left side.
And in case, when the mole of gases on both the sides of the equation i.e. reactant side and product side are equal, it will not have any effect on the equilibrium position on increasing or decreasing volume, or pressure.
I hope my calculation is correct :P
