Answer:
Molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:
Given;
0.60 M concentration of NaOH contains 2.0 L
3.0 M concentration of NaOH contains 495 mL
Molarity is given as concentration of the solute per liters of the solvent.
If the volumes of the two solutions are additive, then;
the total volume of NaOH = 2 L + 0.495 L = 2.495 L
the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M
Molarity of NaOH solution = 3.6 / 2.495
Molarity of NaOH solution = 1.443 M/L
Therefore, molarity of the sodium hydroxide solution is 1.443 M/L
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!
They all have the same number of electrons.