Answer:
c) 0.080 M Al₂(SO₄)₃
Explanation:
Ion [SO₄²⁻] concentration of each solution is:
a) 0.075 M H₂SO₄: <em>[SO₄²⁻] = 0.075M</em>. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻
b) 0.15 M Na₂SO₄: <em>[SO₄²⁻] = 0.15M</em>. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻
c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 =<em> 0.240M</em>. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.
<h3>Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃</h3>
Answer:
The energy that flows into the Earth system is equal to the energy that flows out of the Earth system.
Explanation:
Because I geussed it right
Answer:
Radiation is being released from the reactor.
Explanation:
( A P E X )
4 mol NH₃ → 5 mol N₂
x mol NH₃ → 0.824 mol N₂
x=0.824*4/5=0.6592 mol
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
