In a given sample, hydrogen gas effuses four times faster than the other gas present in the mixture. What is the molar mass and
identity of the unknown gas if the molar mass of hydrogen is 2 grams? 16 grams and oxygen 32 grams and oxygen 8 grams and nitrogen 4 grams and helium 64 grams and ozone
1 answer:
We need to use the effusion formula. I have attached a picture of what this formula looks like because it is a bit hard to type.
effusion formula---> Rate 1/ rate 2= √(M2/M1)
as the questions states, hydrogen gas rate is 4 times the rate of the other gas. so, let call gas 1 the unknown and gas 2 the hydrogen
Rate 1= 1
Rate 2= 4
M1=??
M2= 2 grams
now let's plug in the values and solve for the unknown
1/4 = √(2/x)
x= 32 grams
with the choices given, the gases are either O₂, N₂ or O₃.
O₂ is the one whose mass is 32 because oxygen atomic mass in the periodic table is 16 grams. since we have two oxygens, then 16 x 2= 32
the answer is "<span>32 grams and oxygen "</span>
effusion formula---> Rate 1/ rate 2= √(M2/M1)
as the questions states, hydrogen gas rate is 4 times the rate of the other gas. so, let call gas 1 the unknown and gas 2 the hydrogen
Rate 1= 1
Rate 2= 4
M1=??
M2= 2 grams
now let's plug in the values and solve for the unknown
1/4 = √(2/x)
x= 32 grams
with the choices given, the gases are either O₂, N₂ or O₃.
O₂ is the one whose mass is 32 because oxygen atomic mass in the periodic table is 16 grams. since we have two oxygens, then 16 x 2= 32
the answer is "<span>32 grams and oxygen "</span>
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Answer:
0.83Cal
Explanation:
Given parameters:
Amount of energy given;
35J;
Convert to Cal;
1 calorie = 42 joules
x calories = 35 joules
Cross multiply;
42x = 35
x = = 0.83Cal