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Svetlanka [38]
3 years ago
14

If the same large amount of heat is added to a 250 g piece of aluminum and a 150 g piece of aluminum, what will happen?

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

The 150 g Al will reach a higher temperature.

Explanation:

  • The amount of heat added to a substance (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat added,

m is the mass of the substance,

c is the specific heat of the substance,

ΔT is the temperature difference (final T - initial T).

  • Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).

∵ ΔT is inversely proportional to the mass of the substance.

∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).

<em>So, the right choice is: The 150 g Al will reach a higher temperature.</em>

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After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre
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Answer:

\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}

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Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}

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The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

                        Ni²⁺             +             6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:    6.106×10⁻³                     0.1462                       0

C/mol·L⁻¹:  -6.106×10⁻³         0.1462-6×6.106×10⁻³             0

E/mol·L⁻¹:           0                              0.1095                6.106×10⁻³

Then we approach equilibrium from the right.

                            Ni²⁺   +   6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:              0           0.1095                6.106×10⁻³

C/mol·L⁻¹:            +x            +6x                           -x

E/mol·L⁻¹:             x         0.1095+6x            6.106×10⁻³-x

K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}

Kf is large, so x ≪ 6.106×10⁻³. Then

K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}

 

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