Answer:
77.58g/mol
Explanation:
Let the rate of effusion of NO2 be R1
The rate of effusion of the unknown gas(R2) = 0.770R1
Molar Mass of NO2 (M1) = 14 + (2x16) = 14 + 32 = 46g/mol
The molar mass of the unknown gas (M2) =?
R1/R2 = √(M2/M1)
R1/0.770R1 = √(M2/46)
1/0.770 = √(M2/46)
Take the square of both sides
(1/0.770)^2 = M2/46
Cross multiply to express in linear form
M2 = (1/0.770)^2 x 46
M2 = 77.58g/mol
The molar mass of the unknown gas is 77.58g/mol
Answer:
It is the second leading cause of lung cancer in the USA
The flowers bloom before the leaves grow because the sun is able to get to the forest floor. When the leaves grow, they take up all the sunlight and cast darkness on the forest floor, so the wildflowers can’t bloom anymore. Without sunlight, they can’t photosynthesize and create sugar/food for themselves.
The problem tells us that the equation is already balanced, so we know that we don't have to change any of the coefficients in front of the compounds. Therefore, we can find the ratio by looking at the coefficients of NO and NO²:
2NO : 2NO²
This can be simplified to 1 : 1. Therefore, it is a 1 to 1 ratio.
Answer:
Mass = 17.12 g
Explanation:
Given data:
Mass of Al = 3.90 g
Mass of H₂SO₄ = 13.65
Mass of aluminium sulfate = ?
Solution:
Chemical equation:
3H₂SO₄ + 2Al → Al₂(SO₄)₃ + 3H₂
Now we will calculate the number of moles of each reactant.
Moles of H₂SO₄:
Number of moles = mass/ molar mass
Number of moles = 13.65 g/ 98.079 g/mol
Number of moles = 0.14 mol
Moles of Al:
Number of moles = mass/ molar mass
Number of moles = 3.90 g/ 27 g/mol
Number of moles = 0.14 mol
Now we will compare the moles of aluminium sulfate with sulfuric acid and aluminium.
H₂SO₄ : Al₂(SO₄)₃
3 : 1
0.14 : 1/3×0.14 = 0.05
Al : Al₂(SO₄)₃
2 : 1
0.14 : 1/2×0.14 = 0.07
The number of moles of aluminium sulfate produced by sulfuric acid are less so it will limiting reactant and limit the amount of aluminium sulfate.
Mass of aluminium sulfate:
Mass = number of moles × molar mass
Mass = 0.05 mol × 342.15 g/mol
Mass = 17.12 g