Answer:
You must add 48.97 mL of water to make the 0.200 M diluted solution.
Explanation:
In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.
In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.
The expression for the dilution calculations is:
Cinitial* Vinitial = Cfinal* Vfinal
In this case:
- Cinitial= 12 M
- Vinitial= 0.830 mL
- Cfinal= 0.200 M
- Vfinal= ?
Replacing:
12 M*0.830 mL= 0.200 M*Vfinal
Solving:
Vfinal= 49.8 mL
Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:
49.8 mL - 0.830 mL= 48.97 mL
<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>
The answer is NaCl
-Hope this helps
W*V i believe because it comes down to this :
I is the current, W is wattage V is volts and the * is the thing that represents the Amperes (i’m not 100% but this is my best)
Answer:
0.135 mole of H2.
Explanation:
We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:
Mass of Mg = 3.24 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 3.24/24
Mole of Mg = 0.135 mole
Next, we shall write the balanced equation for the reaction. This is illustrated below:
Mg + 2HCl —> MgCl2 + H2
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.
Thus, 0.135 mole of H2 can be obtained from the reaction.
