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3 years ago

Suppose you have an aqueous solution containing 158.2 g KOHper

Chemistry

1 answer:

postnew [5]3 years ago

4 0

Answer:

1132.8 ml of water

Explanation:

you have an aqueous solution contains 158.2 g KOH per liter

so concentration =158.2/56 = 2.825M

Molarity =2.825

that means you have 2.825 moles of KOH in 1.00L solution

Mass of Soluet(KOH)= 152.8g

Volume of solution= 1.00L

density of solution= 1.13g/cm3 =1.13g/ml

therefore mass of solution = VolumeX density = 1000mL X 1.13g/ml.=1130g

Mass of solvent(water)= mass of solution- mass of solute(KOH)=1130-152.8= 997.2g

Molality= moles of solute/mass of solvent(Kg)

=2.825/(997.2/1000)= 2.832molal

to prepare a 0.250 molal solution of KOH, starting with 100.0ml ofthe orginal solution

0.250*X =2.832 *100

X = 1132.8 ml of water you have to add

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Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

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Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

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3 years ago

A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

#SPJ4

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1 year ago