The concentration of the solution reduces and the number of moles of solute isn't affected.
Data;
- V1 = 50mL
- C1 = 12.0M
- V2 = 200mL
- C2 = ?
<h3>Facts about the diluted solution</h3>
1. When the solution is diluted, the concentration changes and this time, the concentration reduces.
Using dilution formula
The concentration of the solution reduces.
2. The number of moles remains the same.
When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.
Learn more on concentration of a solution here;
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That law is known as Boyle's Law, "The volume of a given mass of a gas is inversely related to pressure when the temperature is constant"
The heat released by the water when it cools down by a temperature difference AT
is Q = mC,AT
where
m=432 g is the mass of the water
C, = 4.18J/gºC
is the specific heat capacity of water
AT = 71°C -18°C = 530
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J
and this is the amount of heat released by the water.
From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637