<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L
<u>Explanation:</u>
We are given:
Volume of mixture = 16 L
Percent of acid present = 30 %
Calculating the percentage of acid present in the mixture:
The mixture is made entirely of acid and water.
Volume of acid in the mixture = 4.8 L
Volume of water in the mixture = 16 - 4.8 = 11.2 L
Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L
Answer:
13.4g
Explanation:
we know that:
1 mole = 6.02 × 10²³ atoms
make the unknown number of moles = x
x = 7.1 × 10²² atoms
putting them both together:
1 mole = 6.02 × 10²³ atoms
x = 7.1 × 10²² atoms
Cross multiply:
6.02 × 10²³ x = 7.1 × 10²²
divide both sides by 6.02 × 10²³
we now have the number of moles of Al₂CO₃
to calculate the grams (mass):
add up all of the atomic masses of Al₂CO₃ to calculate relative formula mass:
(27 × 2) + 12 + (16 × 3) = 114
the grams (mass) of Al₂CO₃:
Answer:
n = 1.7 moles
Explanation:
Given mass, m = 23.8 grams
Molar mass of Nitrogen = 14 g/mol
Let there is n number of moles. We know that,
No. of moles = given mass/ molar mass
So, there are 1.7 moles in 23.8 grams of Nitrogen.