A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final
2 answers:
Answer : The initial temperature of the limestone is .
Solution : Given,
Mass of limestone = 62.6 g
Mass of water = 75 g
Final temperature of limestone =
Final temperature of water =
Initial temperature of water =
The specific heat capacity of limestone =
The specific heat capacity of water =
The formula used :
q = m × c × ΔT
where,
q = heat required
m = mass of an element
c = heat capacity
ΔT = change in temperature
According to this, the energy as heat lost by the system is equal to the gained by the surroundings.
Now the above formula converted and we get
Now put all the given values in this formula, we get
By rearranging the terms, we get the value of initial temperature of limestone.
Therefore, the initial temperature of the limestone is .
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Answer:
Approximately .
Explanation:
Initial volume of the solution: .
Initial quantity of :
.
Ammonia reacts with hydrochloric
acid at a one-to-one ratio:
.
Hence, approximately of
molecules would be required to exactly react with the
in the original solution and hence reach the equivalence point of this titration.
Calculate the volume of that
solution required for reaching the equivalence point of this titration:
.
Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately .
If no hydrolysis took place, of
would be produced. Because
is a soluble salt, the solution would contain
of
ions. The concentration of
would be approximately:
.
However, because is a weak base, its conjugate
would be a weak base.
.
Hence, the following reversible reaction would be take place in the solution at the equivalence point:
.
Let be the increase in the concentration of
in this solution because of this reversible reaction. (Notice that
.) Construct the following
table:
.
Thus, at equilibrium:
- Concentration of the weak acid:
.
- Concentration of the conjugate of the weak acid:
.
- Concentration of
:
.
.
Solve for . (Notice that the value of
is likely to be much smaller than
. Hence, the denominator on the left-hand side
.)
.
Hence, the concentration of at the equivalence point of this titration would be approximately
.
Hence, the at the equivalence point of this titration would be:
.
Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law: