Answer:
34.02 g.
Explanation:
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In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:

So probably, the gas may be H₂S.
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Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)
Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g
Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.
Answer:
3.1atm
Explanation:
Given parameters:
Volume of gas = 2L
Number of moles = 0.25mol
Temperature = 25°C = 25 + 273 = 298K
Unknown:
Pressure of the gas = ?
Solution:
To solve this problem, we use the ideal gas equation.
This is given as;
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
T is the temperature
P =
Now insert the parameters and solve;
P =
= 3.1atm
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100
The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams
The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams