Answer:
10.6%
Explanation:
The determined percent mass of water can be calculated from the formula of the hydrate by
dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiplying this fraction by 100.
Manganese(ii) sulphate monohydrate is MnSO4 . H2O
1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of
hydration must be included.
1 Manganes 52.94 g = 63.55 g
1 Sulphur 32.07 g =
32.07 g 2 Hydrogen is = 2.02 g
4 Oygen =
64.00 g 1 Oxygen 16.00 = 16.00 g
151.01 g/mol 18.02 g/mol
Formula Mass = 151.01 + (18.02) = 169.03 g/mol
2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiply this fraction by 100.
Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%
The final result is 10.6% after the two steps calculations
The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
What is percentage mass?The percentage mass is the ratio of the mass of the element or molecule in the given compound.
The percentage can be given as:
[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]
The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.
Thus,
[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]
Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
Learn more about percentage mass:
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A chemist adds of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.
The question is incomplete, here is the complete question:
A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.
Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The mass of copper (II) fluoride is 0.13 mg
Explanation:
We are given:
Millimolarity of copper (II) fluoride = 0.0013 mM
This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution
Converting millimoles into moles, we use the conversion factor:
1 moles = 1000 millimoles
So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]
Molar mass of copper (II) fluoride = 101.5 g/mol
Putting values in above equation, we get:
[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]
Converting this into milligrams, we use the conversion factor:
1 g = 1000 mg
So,
[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]
Hence, the mass of copper (II) fluoride is 0.13 mg
Answer:
The mass of copper(II) fluoride is 45.54 micrograms
Explanation:
A chemist adds 345.0 mL of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.
Step 1: Data given
Molarity of the copper(II) fluoride solution = 0.0013 mM =
Volume of the solution 345.0 mL = 0.345 L
Molar mass copper(II) fluoride = 101.54 g/mol
Step 2: Calculate moles of copper(II) fluoride
Moles CuF2 = molarity * volume
Moles CuF2 = 0.0000013 M * 0.345 L
Moles CuF2 = 0.0000004485 moles
Step 3: Calculate mass of CuF2
Mass CuF2 = moles * Molar mass
Mass CuF2 = 0.0000004485 moles * 101.54 g/mol
Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms
The mass of copper(II) fluoride is 45.54 micrograms
Identify the factors that govern the speed and direction of a reaction. Check all that apply.
a. Reaction rates increase when the products are more concentrated
b. Reaction rates increase when the reactants are more concentrated
c. Reaction rates increase as the temperaturenses
d. Reaction rates decrease when als we present
Answer: option B and option C
Explanation: