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3 years ago

how can knowledge of percent composition help you as a consumer ? how can you promote responsible consumerism ?

Chemistry

1 answer:

Nonamiya [84]3 years ago

8 0

Answer:

it is a nice question....my mind tells me that the first is it use me as a good vibes and can use to anything the second i will do my best too absorbe it.

Explanation:

Hope this help...

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

3 0

3 years ago

What are common materials

disa [49]

Steel

Fabric (Cotton, Polyester, Felt)

Plastic

Wood

Paper

3 0

3 years ago

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by= $\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ = $moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield = $\frac{374.7g}{486.6g}\times 100=77.0\%$

3 0

3 years ago

Which of the following correctly describes what happens to the kinetic energy of water when it changes from steam to a liquid?

IgorLugansk [536]

Answer:

kinetic energy decreases

Explanation:

kinetic energy usually increases when hot and decreases when cold. The steam is turning into a liquid because it is cooling down, causing the molecules to move less and slow down.

6 0

3 years ago

A food is initially at a moisture content of 90% dry basis. Calculate the moisture content in wet basis

Anvisha [2.4K]

Answer:

Moisture content in wet basis = 47.4 %

Explanation:

Moisture content expresses the amount of water present in a moist sample. Dry basis and wet basis are widely used to express moisture content.

The next equation express the moisture content in wet basis:

$MC_{wb}=\frac{MC_{db}}{1+MC_{db}}$