Adding and subtracting it's how many are behind the decimal, and multiply its how many digits
Answer is: 4.826 grams of anhydrous nickel(II) sulfate could be obtained.
m(NiSO₄×7H₂O) = 8.753 g; mass of heptahydrate.
m(NiSO₄×6H₂O) = 8.192 g; mass of hexahydrate.
m(H₂O) = m(NiSO₄×7H₂O) - m(NiSO₄×6H₂O).
m(H₂O) = 8.753 g - 8.192 g.
m(H₂O) = 0.561 g.
m(NiSO₄) = m(NiSO₄×7H₂O) - 7 · m(H₂O).
m(NiSO₄) = 8.753 g - 7 · 0.561 g.
m(NiSO₄) = 4.826 g.
Answer:
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Explanation:
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<em>Answer:</em>
- Conc. of K+ ions = 0.90 M
- Coc. of SO4∧-2 = 0.45 M
<em>Explanation:</em>
<em>Data Given:</em>
Conc. of H2SO4 = 0.450
As sulphoric acid is a strong electrolyte, it completely dissociate into ions.
H2SO4 ⇆ 2K+ + SO4∧-2
.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.
K2SO4 : K+ K2SO4 : SO4∧-2
1 = 2 1 = 1
0.450 = 2× 0.450 = 0.90 0.450 = 0.450×1 = 0.450
<em> Result:</em>
Conc. of potassium ion will be 0.90M
Coc. of sulphate ions will be 0.45 M
