Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
The answer you are looking for is True
After careful consideration your answer is...
Leucippus and Democritus
*Hope I helped*
~Alanna~
Answer:
612 K
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 306 K
Initial pressure (P₁) = 150 kPa
Final pressure (P₂) = 300 kPa
Volume = 4 L = constant
Final temperature (T₂) =?
Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
150 / 306 = 300 / T₂
Cross multiply
150 × T₂ = 306 × 300
150 × T₂ = 91800
Divide both side by 150
T₂ = 91800 / 150
T₂ = 612 K
Thus, the new temperature of the gas is 612 K
