Answer:
9.01%
Explanation:
The following data were obtained from the question:
Mass of sodium nitrate, NaNO3 = 56.5g
Mass of solution = 627g
The percentage composition of sodium nitrate, NaNO3 in the solution can be obtained as follow:
Percentage composition of NaNO3 = Mass of NaNO3/mass of solution x 100
Percentage composition of NaNO3 = 56.5/627 x 100 = 9.01%
Therefore, the percentage composition of sodium nitrate, NaNO3 in the solution is 9.01%
Answer:
152.63 ml.
Explanation:
By proportion to get a 4.75 M solution the volume will be
(12/4.75) * 100 = 252.63 ml
So the answer is 152.63 ml.
Answer:
Mass of nitrogen dioxide produced = 4.6 g
Explanation:
Given data:
Volume of ammonia = 2.30 L
Mass of nitrogen dioxide produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Number of moles of ammonia at STP:
PV = nRT
n = PV/RT
n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K
n = 2.30 atm .L / 22.414 atm.L/mol
n = 0.1 mol
Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.
NH₃ : NO₂
4 : 4
0.1 : 0.1
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.1 mol × 46 g/mol
Mass = 4.6 g
Answer:
melting
Explanation:
liquid is a good conductor of electricity
