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
It is an endothermic reaction because the products hAve more heat than the reactions so it was a gain of heat which makes the enthalpy Change positive !
Answer: 1.32
Explanation:
Given
85 gm of sulfur dioxide is present at STP
The molar mass of sulfur dioxide is 
The number of moles of sulfur dioxide is
At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D
