Answer: The total volume of the solution produced is 32.5 L.
Explanation:
Molarity denotes the quantity of solute included in 1 liter of solution. The formula for calculating molarity is as follows:
Given values:
Molarity of the solution = 0.02 mol/L
Mass of NaCl = 38 grams
Molar mass of NaCl = 58.5 g/mol
Substituting these values into the equation results in:
Therefore, the total volume of the solution produced is 32.5 L.
Answer:
The integer value of x in the hydrate is 10.
Explanation:
Molar concentration of the solution = 0.0366 M
Volume of the solution = 5.00 L
Moles of hydrated sodium carbonate = n
Weight of hydrated sodium carbonate = n = 52.2 g
Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol
By solving for x, we arrive at:
x = 9.95, approximating to 10
The integer x in the hydrate equals 10.
Answer:
Explanation:
This scenario is unrealistic since Al(OH)₃ is not soluble in water.
The question consists of two parts:
A. Stoichiometry — where we determine volumes, masses, and moles for the products
B. Calorimetry — where we assess the enthalpy of the reaction.
A. Stoichiometry
1. Determine the volume of Al(OH)₃
(a) The balanced chemical equation:
2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O
M/V: 66.667
c/mol·L⁻¹: 4.000 3.000
(b) Moles of H₂SO₄
(c) Moles of Al(OH)₃
The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄
(d) Volume of Al(OH)₃
B. Calorimetry
This reaction has two energy exchanges.
q₁ = heat from the reaction
q₂ = heat used to heat the calorimeter
q₁ + q₂ = 0
nΔH + mCΔT = 0
Data:
Moles of Al₂(SO₄)₃ = 0.066 667 mol
C = 1.10 J°C⁻¹g⁻¹
T_initial = 22.3 °C
T_final = 24.7 °C
Calculations
(a) Mass of solution
Assume solutions are as dense as water (though not realistic).
Mass of sulfuric acid solution = 66.667 g
Mass of aluminium hydroxide solution = 50.000
TOTAL = 116.667 g
(b) ΔT
ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C
(c) ΔH
This result appears nonsensical, but it is derived from your given figures.