Explanation:
Water can be purified from through...
Filtration - This involves the use of filter paper to remove solid objects found in water.
Boiling - This involves keeping water at 100 deg celcius for a while through application of heat and this kills microorganisms in the water.
Addition of alum - Alum causes very fine and small particles that are suspending in water to stick together making them large enough to be removed through filtration.
Chemicals used in water treatment...
Chlorine - Chlorine is used to kill bacteria, viruses and other disease causing pathogens found in water
That's all I know...
Answer:
See explaination
Explanation:
1) Pb(NO3)2 => Pb2+ + 2 NO3-
[Pb2+] = [Pb(NO3)2] = 7.56 mM = 7.56 x 10-3 M
Pb(IO3)2 <=> Pb2+ + 2 IO3-
Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13
7.56 x 10-3 x [IO3-]2 = 2.5 x 10-13
[IO3-] = 5.75 x 10-6 M ≈ 5.8 x 10-6 M
(2) [Pb2+] = 1.7 x 10-6 M
Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13
1.7 x 10-6 x [IO3-]2 = 2.5 x 10-13
[IO3-] = 3.83 x 10-4 M
[IO3-]from Pb(IO3)2 = 2 x [Pb2+]
= 2 x 1.7 x 10-6 = 3.4 x 10-6 M
[IO3-]from NaIO3 = [IO3-] - [IO3-]from Pb(IO3)2
= 3.83 x 10-4 - 3.4 x 10-6
= 3.80 x 10-4 M
NaIO3 => Na+ + IO3-
[NaIO3] = [IO3-]from NaIO3
= 3.80 x 10-4 M ≈ 3.8 x 10-4 M
Answer:

Explanation:
Hello!
In this case, since the undergoing chemical reaction is:

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

Best regards!
The question is incomplete, the complete question is;
Using the following equation 2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq) how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid
Answer:
355.1 g
Explanation:
The equation of the reaction is;
2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq)
We have been told that H2SO4 is in excess so NaOH is the limiting reactant. Therefore;
Number of moles in 200g of NaOH = 200g/40g/mol = 5 moles
So;
2 moles of NaOH yields 1 mole of Na2SO4
5 moles of NaOH will yield 5 * 1/2 = 2.5 moles of Na2SO4
Molar mass of Na2SO4 = 142.04 g/mol
Mass of Na2SO4= 2.5 moles * 142.04 g/mol = 355.1 g
Starting mass: 294
Mass ratio: 3/5