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3 years ago

Part A

Chemistry

1 answer:

abruzzese [7]3 years ago

6 0

117 mL of 0.210 M K₂S solution

Explanation:

The question asks about the volume of 0.210 M K₂S (potassium sulfide) solution required to completely react with 175 mL of 0.140 M Co(NO₃)₂ (cobalt(II) nitrate).

We have the chemical reaction:

K₂S + Co(NO₃)₂ → CoS + 2 KNO₃

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume

number of moles of Co(NO₃)₂ = 0.140 × 175 = 24.5 mmoles

We see from the chemical reaction that 1 mmole of Co(NO₃)₂ is reacting with 1 mmole of K₂S, so 24.5 mmoles of Co(NO₃)₂ are reacting with 24.5 mmoles of K₂S.

volume = number of moles / molar concentration

volume of K₂S solution = 24.5 / 0.210 = 117 mL

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molar concentration

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Part D

Basile [38]

The heat lost by copper(ii) sulfate is equal to heat absorbed by water since the total energy in the system remains constant according to the law of conservation of energy.

<h3>How can the number of moles be determined?</h3>

The number of moles of a substance is determined using the formula below:

Number of moles = mass/molar mass

Assuming the mass of copper(ii) sulfate used is <em>Mc</em>, number of moles of copper(ii) sulfate used is:

Moles of copper(ii) sulfate = <em>Mc</em>/159.60 moles

The heat absorbed by water is calculated using the formula below:

Quantity of Heat, H = mass × specific heat capacity × temperature change

mass of water <em>=</em><em> </em> 10 g

Let temperature change be <em>Tc</em>

Heat<em> </em>absorbed<em> </em>by water = 10 × 4.186 × Tc = 41

86Tc

The change in internal energy, ΔU of copper(ii) sulfate, is given as:

ΔU = Q − W

where:

Q = heat absorbed by water

W = work done by or on the system

The enthalpy of the reaction is given as:

ΔH= energy released or absorbed/moles of copper (ii) sulfate

Therefore, according to the law of conservation of energy, the total energy in the system remains constant.

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2 years ago

How many grams of oxygen can be prepared by the decomposition of 12 grams of mercury oxide<br><br>

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Taking into account the reaction stoichiometry, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction of the decomposition of mercury oxide is:

2 HgO → 2 Hg + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

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<h3>Mass of oxygen formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 433.18 grams of HgO form 32 grams of O₂, 12 grams of HgO form how much mass of O₂?

$mass of O_{2} =\frac{12 grams of HgOx 32 grams of O_{2}}{433.18 grams of HgO}$

<u><em>mass of O₂= 0.886 grams</em></u>

Then, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

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