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.
Answer:
The answer is Graph A, because there is a direct relationship between pressure and volume.
The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure.
Answer:
Conduction
Explanation:
The handle is touching the pot.
And I got it wrong
The answer is C
explanation
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
