Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
Answer:
Explanation:
We need to use the formula for heat of vaporization.
Identify the variables.
- The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
- Q is the energy, in this problem, 50,000 Joules.
- m is the mass, which is unknown.
Substitute the values into the formula.
We want to find the mass. We must isolate the variable, m.
m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.
Divide. Note that the Joules (J) will cancel each other out.
Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.
The mass is about 22 grams, so choice B is correct.
Answer:
100%
Explanation:
My teacher just asked this question.
From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
From 5 moles to liters, just multiply 5 by 22.4. I got 112 L of H2.
