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Leno4ka [110]
3 years ago
12

Zn2+ and Al3+ are amphoteric. Demonstrate

Chemistry
1 answer:
denis23 [38]3 years ago
8 0

Explanation:

1) Amphoteric substances can react both like an acid or a base depending if they are in presence of a base or an acid.

Zinc hydroxide (Zn(OH)2):

In acid: Zn(OH)_2 + 2 HCl \longrightarrow ZnCl_2 + 2 H_2O


In base: Zn(OH)_2 + 2 NaOH \longrightarrow Na_2[Zn(OH)_4]

Aluminium hydroxide (Al(OH)3)

In acid: Al(OH)_3 + 3 HCl \longrightarrow AlCl_3 + 3 H_2O


In base: Al(OH)_3 + NaOH \longrightarrow Na[Al(OH)_4]

2) Reactions

a) 2 NH_3 + Cu^{2+} + 2 H_2O \longrightarrow 2 NH_4^+ + Cu(OH)_2

3 NH_3 + Fe^{3+} + 3 H_2O \longrightarrow 3 NH_4^+ + Fe(OH)_3

b) 2 NH_3 + Zn^{2+} + 2 H_2O \longrightarrow 2 NH_4^+ + Zn(OH)_2

3 NH_3 + Al^{3+} + 3 H_2O \longrightarrow 3 NH_4^+ + Al(OH)_3

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If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t
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Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
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4 years ago
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If the half-life of 37Rb is 4.7x101 years, how long would it take for 0.5 grams of a 2 gram sample to radioactively decay?
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<u>Answer:</u> The time required will be 19.18 years

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

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We are given:

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k = rate constant  = 0.015yr^{-1}

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Putting values in above equation, we get:

0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs

Hence, the time required will be 19.18 years

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