Answer:
V₂ = 1866.32 mL
Explanation:
Given data:
Initial temperature = 27.8°C (27 + 273.15 K = 300.15 k)
Initial volume = 1500 mL
Final volume = ?
Final temperature = 100.0°C (100.0 + 273.15 K = 373.15 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
1500 mL / 300.15 k = V₂ / 373.15 K
V₂ = 1500 mL× 373.15 K/ 300.15 k
V₂ = 560175 mL. K /300.15 k
V₂ = 1866.32 mL
Answer:
the bike is moving at a speed of 0.115810 mph
Answer:
- <u><em>The volume of CO₂(g) produced at STP when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is 2.24dm</em></u><em><u>³</u></em>
Explanation:
The question is incomplete.
This is the complete question:
<em>Consider the reaction by the following equation:</em>
<em />
<em> C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)</em>
<em />
<em>The volume of CO₂(g) produced at s.t.p when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is ________</em>
<em />
<em>[Molar Volume of gas = 22.4dm³]</em>
<em />
<em> A. 1.12dm³</em>
<em> B. 2.24dm³</em>
<em> C. 3.72dm³</em>
<em> D. 4.48dm³</em>
<h2>Solution</h2>
<u />
<u>1. Write the mole ratio between CO₂(g) and C₂H₄(g)</u>
- 1 mol C₂H₄(g) : 2 mol CO₂(g)
<u>2. Multiply the 0.05 moles of C₂H₄(g) by the mole ratio</u>
- 0.05 mol C₂H₄ × 2 mol CO₂ / 1 mol C₂H₄(g) = 0.10 mol CO₂
<u>3. Convert moles of CO₂ to volume at STP using the molar volume at STP</u>
- 0.10 mol CO₂ × 22.4 dm³ / mol = 2.24 dm³
First, we convert the given amount of energy into joules.
1 kJ = 1000 joules
2.2125 kJ = 2,212.5 Joules
Each kilocalorie contains 4,184 Joules
Kilocalories = 2,212.5 / 4,182
Kilocalories = 0.529
1 kilocalorie = 1000 calories
0.529 kilocalories = 529 calories