When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = =74.8 %.
Answer: most store-bought isnt
Explanation:
there typically a lot of addivitves
Explanation:
The given chemical equation is:
The rate of the reaction is 0.0352 M/s.
During the course of the reaction, the rate of reactants decreases, and the rate of products increases.
The rate of disappearance of B is shown below:
So, rate of change of B is :
Option C.
Answer:
17.6510 L
Explanation:
First we should get the number of moles of helium here by Boyle's law
PV=nRT
P=750/760= 0.9868 atm
T=25+273=298 kelvin
R= 0.08206
V= 20L
so
n=PV/RT
n=0.9868×20/0.08206×298
n=0.80707 mol
Then use the same law
V=0.80707×0.08206×263/0.9869=
17.6510L
SO THE VOLUME WILL BE 17.6510 L