On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Answer:
The concentration after 20 mins is 0.832 M
Explanation:
Zero order rate law is given by;
R = K [A₀]⁰
A zero order reaction, rate is independent of the initial concentration
R = K
Where;
R is the rate of reaction
K is the rate constant = 0.0416 M/min
Since R = K,
Then, R = 0.0416 M/min
After 20 min, the concentration will be;
A = Rt
A = (0.0416 M/min)(20 min)
A = 0.832 M
Therefore, the concentration after 20 mins is 0.832 M
<h2>
Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
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Why?</h2>
To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:
Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:
Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!
to the proper number of significant figures) to the following? (12.67+19.2)(3.99)/(1.36+ 11.366).
Propanoic acid formula is ch ch 2 so it has 8 bonds
