What is the molarity of a 3.0 solution of magnesium chloride prepared with 9.0 molecules of MgCl2 A. 6.0 M B. 3.0 M C. 27.0 M D.
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Answer: 6.
Explanation:
1) Aluminum
So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium
So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:
4) Net equation
Add the two half-equations:
As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
Explanation:
1) Aluminum
So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium
So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:
4) Net equation
Add the two half-equations:
As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ =
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL