What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi
1 answer:
Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂
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Answer:
The volume will be 185.83 mL.
Explanation:
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, k = Constant
Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. Boyle's law is expressed mathematically as:
P*V=k
Where P = pressure, V = volume, k = Constant
Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is kept at a constant pressure. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement of the gas molecules. .
In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
In this case:
- P1= 720 torr
- V1= 256 mL
- T1= 35 C= 308 K (being 0 C= 273 K)
- P2= 1.25 atm= 950 torr (being 1 atm= 760 torr)
- V2= ?
- T2= 22 C= 295 K
Replacing:
Solving:
V2= 185.83 mL
<u><em>The volume will be 185.83 mL.</em></u>