Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
To convert take 1.8 * degrees C + 32
1.8 * 35 + 32
63 + 32 = 95 so C
We can express the rate equation in this form:
-r = k A^n B^m
where -r is the rate
k is the rate constant,
A is the concentration of CH3Cl
n is the order with respect to CH3Cl
B is the concentration of H2O
m is the order with respect to H2O
We can solve this by trial and error or by calculus. The first method is easier. The rate constant does not depend on the concentration of the reactant. Assume values of n and m and solve for k in each experiment. The only option that gives really close values of k in each experiment is:
<span>C. CH3Cl: firstorder H2O: second order
</span>
Answer:
Highest boiling point - 0.43 m Urea
Second highest boiling point - 0.20 m NiSO4
Third highest boiling point - 0.19 m NH4I
Lowest boiling point - 0.17 m NH4NO3
Explanation:
We know that;
ΔT = kb m i
Where;
ΔT = boiling point elevation
kb = boiling point constant
m = molality of the solution
i = Van't Hoff factor
For NiSO4 , NH4I and NH4NO3 , the Van't Hoff factor, i = 2
But for Urea, the Van't Hoff factor, i = 1
We also have to consider both the values of the molality and Van't Hoff factor , knowing that a higher molality and a higher Van't Hoff factor leads to a higher ΔT and consequently a higher boiling point.
This facts above account for the arrangement of substances shown in the answer.
