Answer:
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Explanation:
Moles of copper = 
According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.
Then 0.03613 moles of copper will give:
of nitrogen dioxide gas
Moles of nitrogen dioxide gas = n = 0.06326 mol
Pressure of the gas = P
P = Total pressure - vapor pressure of water
P = 726 mmHg - 23.8 mmHg = 702.2 mmHg
P = 0.924 atm (1 atm = 760 mmHg)
Temperature of the gas = T = 25.0°C =298.15 K
Volume of the gas = V
V = 1.68 L
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Answer:
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the answer is "There is an outside energy source." in voltaic cells, there is not need of an external source of energy because they involve spontaneous reactions.
in both types of cells, the electrons move from anode to cathode, the anode is where oxidation takes place and cathode for reduction (an ox, red cat). also both have two half reactions.
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
