In is a lot easier because it uses 10s.
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions×
= 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄×
= <em>34,6g of (NH₄)₂SO₄</em>
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I hope it helps!
6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.
Answer:
pH= 3.82
Explanation:
Sodium ethanoate or sodium acetate (CH3COONa) ionises completely.
Ethanoic acid or acetic acid (CH3COOH), ionises partially because it is a weak acid.
Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄