What is the molarity of a solution prepared by diluting 43.72 ml of 1.005 M aqueous K2CR2O7 to 500 ml
1 answer:
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Answer:
\large \boxed{\textbf{609 kJ}}
Explanation:
The formula for the heat absorbed is
q = mCΔT
Data:
m = 2.07 kg
T₁ = 23 °C
T₂ = 191 °C
C = 1.75 J·°C⁻¹g⁻¹
Calculations:
1. Convert kilograms to grams
2.07 kg = 2070 g
2. Calculate ΔT
ΔT = T₂ - T₁ = 191 - 23 = 168 °C
3. Calculate q
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant
3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO
From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂
(b) Mass of NO reacted
(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO