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postnew [5]
3 years ago
14

1- Alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2. XH2O. To find the value of X, you can heat the sample

of compound. Assume you heat 4.74 g of the hydrated compound and that sample loses 2.16 g of water. What is the value of X
Chemistry
1 answer:
Burka [1]3 years ago
7 0

Answer:

THE VALUE OF X IS 7 AND THE FORMULA OF THE HYDRATED SALT IS KAl(SO4)2.7H20

Explanation:

1. write out the varibales given in thequestion:

Mass of the hydrated salt = 4.74 g

Mass of water lost = 2.16 g

Formula of the hydrated salt = KAl(SO4)2. XH20

2. calculate the molar mass of the salt and that of water of crystallization:

Molar mass of anhydrous salt = ( K = 39, Al = 27, S = 32, 0=16)

= ( 39 + 27 + 32*2 + 16 * 8

= (39 + 27 + 64 + 128)

= 258 g/mol

Molar mass of water = 18 g/mol

3. Use this expression to calculate X:

The expression,

XH20 / molar mass of anhydrous salt = Mass of water lost / Mass of hydrated salt.

X = molar mass of anhydrous salt * mass of water lost / mass of anhydrous salt * H20

where XH20 is the molar mass of water of crystallization, is used to calculate the value of X.

4. Solve for X:

So therefore:

X = 258 * 2.16 / 4.74 * 18

X = 557.28/ 85.32

X = 6.53

X is approximately 7.

The value of X is 7 and the formula pf the hydrated salt is KAl(SO4)2.7H20

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Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess
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Answer:

a) yes, it was an hydrate

b) the number of waters of hydration, x = 6

Explanation:

a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.

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mass if dehydrated NiCl2 = 2.3921 grams

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NiCl2.xH2O

mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole

mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole

Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each

for NiCl2 = 0.01846/0.01846 = 1

for H2O = 0.11072/0.01846 = 5.9976 = 6

thus the hydrated sample was NiCl2. 6H2O

4 0
3 years ago
For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re
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Answer:

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In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

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The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

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In a balanced ionic equation:

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Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

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2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

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