Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
The correct name of the compound Mn3(PO4)2 is definitely the last option represented above <span>D. manganese(II) phosphate. I am pretty sure this answer will help you
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Answer:
semipermeability
Explanation:
partially but not freely or wholly permeable specifically : permeable to some usually small molecules but not to other usually larger particles a semipermeable membrane.
Th actual yield of the reaction is 24.86 g
We'll begin by calculating the theoretical yield of the reaction.
2Na + Cl₂ → 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl from the balanced = 2 × 58.5 = 117 g
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore,
11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.
Thus, the theoretical yield of NaCl is 29.25 g.
Finally, we shall determine the actual yield of NaCl.
- Theoretical yield = 29.25 g
Actual yield = Percent yield × Theoretical yield
Actual yield = 85% × 29.25
Actual yield = 0.85 × 29.25 g
Actual yield = 24.86 g
Learn more about stoichiometry: brainly.com/question/25899385
Thus, there are 1.47 × 10^(23) molecules present in 122 grams of NO2.
