Answer:
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Explanation:
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
Answer:
B. The pressure has increased
Explanation:
Answer:
Molality of the solution = 0.7294 M
Explanation:
Given:
Number of magnesium arsenate = 1.24 moles
Mass of solution = 1.74 kg
Find:
Molality of the solution
Computation:
Molality of the solution = Mole of solute / Mass of solution = 1.74 kg
Molality of the solution = 1.24 / 1.7
Molality of the solution = 0.7294 M
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