Answer:
MnSO₄.7H₂O
Explanation:
To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:
<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>
17.51g * (1mol / 151g) = 0.116 moles
<em>Moles H₂O -Molar mass: 18g/mol-:</em>
32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles
The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:
0.813mol / 0.116mol = 7 molecules of water.
The hydrate formula is:
<h3>MnSO₄.7H₂O</h3>
Is soluble in water but not soluble in acetonitrile.
So i think its false
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
Answer:
0.203 is the mean of the concentration of the HCl solution
Explanation:
You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:
0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =
<h3>0.203 is the mean of the concentration of the HCl solution</h3>
