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3 years ago

Which compound does not form a precipitate when reacted with sodium hydroxide?

Chemistry

1 answer:

Alla [95]3 years ago

5 0

Answer:

NaNO₃

Explanation:

A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.

From the choices given precipitation reaction will occur between;

Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)

Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.

From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.

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Answer:

Because iodine is a liquid, it has no melting point. Iodine is not an electrical conductor because each molecule consists of two iodine atoms connected by a covalent bond that cannot be stimulated sufficiently to transmit electrical energy.

Explanation:

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Answer:

3'' one is correct answer

Explanation:

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3 years ago

Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17

Hitman42 [59]

Answer is: mass of burned butane is 11.6 g.

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

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n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
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n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
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3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

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2 years ago

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