Answer:
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Explanation:
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Answer:
T2 = 260 K
Explanation:
<em>Given data:</em>
P1 = 150.0 k Pa
T1 = (-23+ 273.15) K = 250.15 K
V1 = 1.75 L
P2 = 210.0 kPa
V2 = 1.30 L
<em>To find:</em>
T2 = ?
<em>Formula:</em>
<em>Calculation:</em>
T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)
T2 = 260 K
The correct option is C.
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Answer:
The volume of CO2 produced is 6.0 L (option D)
Explanation:
Step 1: Data given
Volume of oxygen = 3.0 L
Carbon monoxide = CO = in excess
Step 2: The balanced equation
2 CO (g) + O2 (g) → 2 CO2 (g)
Step 3: Calculate moles of O2
1 mol of gas at STP = 22.4 L
3.0 L = 0.134 moles
Step 3: Calculate moles of CO2
For 2 moles CO we need 1 mol of O2 to produce 2 moles of CO2
For 0.134 moles O2 we'll have 2*0.134 = 0.268 moles CO2
Step 4: Calculate volume of CO2
1 mol = 22.4 L
0.268 mol = 22.4 * 0.268 = 6.0 L
The volume of CO2 produced is 6.0 L
