The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
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When products form in an exergonic reaction, the <span>energy required to break bonds is less than the energy released from new bonds. This is usually given off as light. I hope this is the answer you are looking for. Looking forward to help you again. Have a great day ahead!</span>
The correct answer is <span>Fusion Curve</span>
Answer:
Percentage yield is 41.21%
Explanation:
Equation of reaction,
N₂ + 3H₂ → 2NH₃
Actual NH3 = 6.83g
Mass of N2 = 5.77g
Theoretical yield = ?
5.77g of N2 = 6.83g of NH3
14g of N2 = xg
X = (14 × 6.83) / 5.77
X = 95.62 / 5.77
X = 16.57g of NH3
Theoretical yield of NH3 is 16.57g
Percentage yield = (actual yield / theoretical yield) × 100
% yield = (6.83 / 16.57) × 100
% yield = 0.4121 × 100
% yield = 41.21%
The percentage yield of NH3 is 41.21%
