They both involve Heat, Oxygen, and Light.
The percentage of water of crystallization in blue vitriol is 36.07%.
M(H₂O) = 2Ar(H) + Ar(O) x g/mol
M(H₂O) = 2 + 16 x g/mol
M(H₂O) = 18 g/mol; molar mass of water
M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol
M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol
M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate
The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%
The percentage of water: 90 g/mol / 249.5 g/mol x 100%
The percentage of water = 36.07%
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Answer:
0.93 mol
Explanation:
Given data:
Number of moles of Na atom = ?
Number of atoms = 5.60× 10²³
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
5.60× 10²³ atoms × 1 mol / 6.022 × 10²³ atoms
0.93 mol
A carbon iota can bond with four other iotas and is just like the four-hole wheel, whereas an oxygen iota, which can bond only to two, is just like the two-hole wheel.
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
