Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
1. Theoretical yield = 2.03g
2. Actual yield 1.89g
Explanation:
Let us write a balanced equation. This is illustrated below:
Zn + 2HCI —> ZnCl2 + H2
Molar Mass of HCl = 1 +35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 2 x 36.5 = 73g
Molar Mass of H2 = 2x1 = 2g/mol
1. From the equation,
73g of HCl produced 2g of H2.
Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g
Therefore, theoretical yield = 2.03g
2. %yield = 93%
Theoretical yield = 2.03g
Actual yield =?
%yield = Actual yield /Theoretical yield x100
Actual yield = %yield x theoretical yield
Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g
Actual yield =1.89g
