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Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Answer:
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Explanation:

Moles of copper = 
According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.
Then 0.03613 moles of copper will give:
of nitrogen dioxide gas
Moles of nitrogen dioxide gas = n = 0.06326 mol
Pressure of the gas = P
P = Total pressure - vapor pressure of water
P = 726 mmHg - 23.8 mmHg = 702.2 mmHg
P = 0.924 atm (1 atm = 760 mmHg)
Temperature of the gas = T = 25.0°C =298.15 K
Volume of the gas = V


V = 1.68 L
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.