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3 years ago

Plzzz help do your best to answer these questions

Chemistry

1 answer:

vovikov84 [41]3 years ago

5 0

The empirical formula is the simplest formula of a chemical compound.

To find the empirical formula, we take the following steps;

Divide the percentage by mass of each element by its relative atomic mass.
Divide the quotient of each by the lowest value obtained instep 1 above
Write the result of step 2 above as the subscript following each atom.

1) O - 88.10/16, H - 11.190/1

O - 5.5, H - 11.19

O - 5.5/5.5, H - 11.19/5.5

O - 1, H - 2

Empirical formula = OH2

2) C - 41.368/12 H - 8.101/1, N - 32.162/14, O - 18.369/16

C - 3, H - 8, N - 2, O - 1

C - 3/1, H - 8/1 N - 2/1 O - 1/1

C - 3, H - 8, N - 2, O - 1

Empirical formula = C3H8N2O

To obtain the molecular formula where n = number of atoms of each element;

Molecular weight = 174.204 g/mol

[ 3(12) + 8(1) + 2(14) + 16]n = 174

n= 174/88

n = 2 (to the nearest whole number)

Hence, we have;

[C3H8N2O]2

The molecular formula is C6H16N4O2

3) C - 19.999/12, H - 6.713/1, N - 46.646/14, O - 26.641/16

C - 2, H - 7, N - 3, O - 2

C - 2/2, H - 7/2, N - 3/2, O - 2/2

C - 1, H - 4, N - 2, O - 1

Empirical formula - CH4N2O

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

Explanation:

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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

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(c) [NH₃]

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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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