Answer:
1.58 M
Explanation:
is 1.66 m concentration.
Which means that 1.66 moles of are present in 1 kg of the solvent, water.
Mass of water = 1 kg = 1000 g
Moles of = 1.66 moles
Molar mass of = 98.079 g/mol
The formula for the calculation of moles is shown below:
Thus,
Total mass = 1000 g + 162.81114 g = 1162.81114 g
Density = 1.104 g/mL
Volume of the solution = Mass / Density = 1162.81114 / 1.104 mL = 1053.27 mL = 1.05327 L
Considering:-
<u>Molarity = moles/ Volume of solution = 1.66 / 1.05327 M = 1.58 M
</u>
The compound ch3ch2-sh is in the organic family know as c. thiols
Answer:
THE MOLARITY OF THE SOLUTION IS 3.36 MOLE/L
Explanation:
First we must understand what molarity is.
Molarity is the number of mole per unit volume of solution. In this question, 9.25 mole of H2SO4 was given in 2.75 L of solution.
Molarity is written in mole per dm3 or L.
So we can calculate the molarity:
9.25 ole of H2SO4 = 2.75 L of solution
The number of mole in 1 L of solution will be:
= 9.25 mole / 2.75 L
= 3.3636 mole/ L
In conclusion, the molarity of the solution is approximately 3.36 mole/L
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
