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Lelechka [254]
3 years ago
13

Explain the difference betweeen mass and volume

Chemistry
1 answer:
sattari [20]3 years ago
7 0
mass is the amount of matter in an object, and volume is the amount of space something takes up
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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
Marina CMI [18]

Answer:

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

Explanation:

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}

2. Moles of NH₃

n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}

(c) [NH₃]

c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}

3. Equilibrium concentration of Ni²⁺

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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Answer:

The answer is in the explanation.

Explanation:

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Based on the experiment, it is possible that the solution Z was a buffer and Y another kind of solution. For this reson, pH of the solution Y changes much more than the pH of solution Z changes despite the amount of NaOH added is the same in both solutions.

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kiruha [24]

60mol O2 × 1 mol C6H12O6 / 6 moles 02 = 10 moles of Glucose

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4 years ago
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