Answer:
1.5 × 10² mL
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 1.9 atm
- Initial volume of the gas (V₁): 80 mL
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final volume of the gas (V₂): ?
Step 2: Calculate the final volume of the gas
For an ideal gas, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1.9 atm × 80 mL/1.0 atm
V₂ = 1.5 × 10² mL
Since the pressure decreased, the volume of the gas increased.
Answer:
The heat released by the combustion is 20,47 kJ
Explanation:
Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:
Q = C×m×ΔT + Cc×ΔT
Where:
Q is the heat released
C is specific heat of water (4,186kJ/kg°C)
m is mass of water (1,00kg)
ΔT is temperature change (23,65°C - 20,45°C)
And Cc is heat capacity of the calorimeter (2,21kJ/°C)
Replacing these values the heat released by the combustion is:
<em>Q = 20,47 kJ</em>
Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?
- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
