Answer : The final temperature of the solution in the calorimeter is, 
Explanation :
First we have to calculate the heat produced.
where,
= enthalpy change = -44.5 kJ/mol
q = heat released = ?
m = mass of 
= 1.52 g
Molar mass of = 40 g/mol
Now put all the given values in the above formula, we get:
Now we have to calculate the final temperature of solution in the calorimeter.
where,
q = heat produced = 1.691 kJ = 1691 J
m = mass of solution = 1.52 + 35.5 = 37.02 g
c = specific heat capacity of water = 
= initial temperature = 
= final temperature = ?
Now put all the given values in the above formula, we get:
Thus, the final temperature of the solution in the calorimeter is,
The relation of heat (Q), mass (m), and change of temperature (ΔT) is given by the formula:
Q = m*Cs*ΔT, where Cs is the specific heat of the material.
Then, given than you know Q, m and ΔT, you can solv for Cs:
Cs = Q / (m*ΔT)
Cs = 89.6 J / [20 grams * (40.0°C - 30.0°C)] = 0.448 J / g * °C.
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
<u>Answer:</u> The equilibrium constant for 
equation is 
<u>Explanation:</u>
The given chemical equation follows:
The value of equilibrium constant for the above equation is 
Calculating the equilibrium constant for the given equation:
The value of equilibrium constant for the above equation will be:
Hence, the equilibrium constant for equation is
