Name the follwing compound, KBr
1 answer:
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Answer:
i
ii
Explanation:
From the question we are told that
The first temperatures is
The second temperature is
Generally the equation for the most highly populated rotational energy level is mathematically represented as
Here R is the gas constant with value
Also
B is given as
Generally the energy require per mole to move 1 cm is 12 J /mole
So will require x J/mole
=>
So at the first temperature
=>
So at the second temperature
=>
Answer:
36.37% is the percent yield of the reaction.
Explanation:
1)0.650 L nitrogen gas , at 295 K and 1.01 bar.
Let the moles of nitrogen gas be n.
Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)
Temperature of the gas = T = 295 K
Volume of the gas = V = 0.650 L
Using an ideal gas equation:
2) Moles of ammonia gas=
Moles of oxygen gas =
According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.
Then,0.1101 mol of oxygen will react with:
of ammonia.
Hence, oxygen gas is in limiting amount and act as limiting reagent.
3) Theoretical yield of nitrogen gas :
According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.
Then 0.1101 mol of oxygen will give:
of nitrogen.
Theoretical yield of nitrogen gas = 0.0734 mol
Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol
Percentage yield:
Percentage yield of the reaction:
36.37% is the percent yield of the reaction.