% yield = 80.719
<h3>Further explanation</h3>
Given
22.0 g of Mgl₂
25.0 g of Mg
25.0 g of l₂
Required
The percent yield
Solution
Reaction
Mg + I₂⇒ MgI₂
mol Mg = 25 g : 24.305 g/mol = 1.029
mol I₂ = 25 g : 253.809 g/mol = 0.098
Limiting reactant = I₂
Excess reactant = Mg
mol MgI₂ based on I₂, so mol MgI₂ = 0.098
Mass MgI₂ (theoretical):
= mol x MW
= 0.098 x 278.114
= 27.255 g
% yield = (actual/theoretical) x 100%
% yield = (22 / 27.255) x 100%
% yield = 80.719
Options:
monoglycerides
cocamide DEA
folic acid
iron chromium ion
peroxide
lauryl glucoside
disodium phosphate
Answer and Explanation:
The added chemicals are:
- monoglycerides
- folic acid
- iron
- disodium phophates
Monoglycerides are fats added for flavour. Folic cid and iron are vitamins added for nutritional value. disodium phosphate is a food additive for enhancing flavour.
The remaining ingredients are organic based.
Answer: The molecular formula will be 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H =
For O =
The ratio of C : H: O= 4: 4:1
Hence the empirical formula is 
The empirical weight of = 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now we have to calculate the molecular formula.
The molecular formula will be=
m = given mass of gas = 3.82 g
M = molar mass of gas = ?
T = temperature of laboratory = 302 K
P = air pressure = 1.04 atm = 1.04 x 101325 pa
V = volume of gas = 0.854 L = 0.854 x 10⁻³ m³
using the ideal gas equation
PV = (m/M) RT
inserting the above values
(1.04 x 101325) (0.854 x 10⁻³) = (3.82/M) (8.314) (302)
M = 106.6 g
hence the molar mass of the gas comes out to be 106.6 g
<span>We know that density is equal to mass divided by volum, D=M/V and in this case we have 1 gallon of a solution of sulfuric acid with 37.4% of concentration in mass.
1 gallon is 3785.41 ml and according the formula M=D*V = 1.31 * 3785.41 = 4958.89 grams of solution.
Only 37.4% of the solution is sulfuric acid, that is 4958.89 * 37.4/100= 1854.62 grams
Then the number of grams of sulfuric acid is 1854.62 gr.</span>
