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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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4 years ago
Calculate the volume occupied by 55.5 moles of water at STP.
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Answer:

Volume occupied by 55.5 moles of water gas at STP = 1240 Liters.

Explanation:

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So, if 1 molar volume of water gas (steam) occupies 22.4 Liters at STP, then 55.5 moles of water gas occupies 22.4 Liters/mole x 55.5 moles = 1243.2 Liters, or 1240 Liters (3 sig. figs.).   :-)

4 0
3 years ago
1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t
slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄  

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄  in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ?            ; c₂ = 2.0   mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500  mL of water.

3 0
4 years ago
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moles of solute= 0.01 mol

molality= 0.01 mol / 0.285 kg
molality= 0.04 m
6 0
3 years ago
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Then bases have pH above 7.

Examples of bases are calcium carbonate, sodium hydroxide, etc.

Thus, the correct option is C.

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