1. Write the empirical and graphic formulas of nitric acid and aluminium
1 answer:
1. HNO₃ nitric acid
Al(NO₃)₃ aluminium nitrate
You may find the the attached picture the graphical representation of both compounds.
2. You may find bellow the full equations
Explanation:
1. HNO₃ nitric acid, Al(NO₃)₃ aluminium nitrate
dissociation equation
HNO₃ (l) + H₂O (l) → NO₃⁻ (aq) + H₃O⁺ (aq)
Al(NO₃)₃ (s) + H₂O (l) → Al³⁺ (aq) + 3 NO₃⁻ (aq)
2. molecular equations
Al(OH)₃ (s) + NaOH (aq) → NaAl(OH)₄ (s)
Ba(OH)₂ (aq) + NiCl₂ (aq) → BaCl₂ (aq) + Ni(OH)₂ (s)
ionic equations
Al(OH)₃ (s) + Na⁺ (aq) + OH⁻ (aq) → NaAl(OH)₄ (s)
Ba²⁺ (aq) + 2 OH⁻ (aq) + Ni²⁺ (aq) + 2 Cl⁻ (aq) → Ba²⁺ (aq) + 2 Cl⁻ (aq) + Ni(OH)₂ (s)
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Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>
I hope it helps!