Answer:
The rate decreases
Explanation:
When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.
Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.
Hence increase in temperature decreases the rate of solubility of a gas in water.
The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution.
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
<span>.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
</span><span>.8x = 14
</span><span>x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)
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Answer:
Q = 1455.12 Joules.
Explanation:
Given the following data;
Mass = 300 grams
Initial temperature = 22.3
Final temperature = 59.9°C
Specific heat capacity = 0.129 J/gºC.
To find the quantity of energy;
Where,
Q represents the heat capacity.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt represents the change in temperature.
dt = T2 - T1
dt = 59.9 - 22.3
dt = 37.6°C
Substituting the values into the equation, we have;
Q = 1455.12 Joules.