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3 years ago

A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Chemistry

2 answers:

LuckyWell [14K]3 years ago

6 0

The hydration number "n" in Sr(NO3)2•nH2O is 4.

<h2>Further Explanation:</h2><h3>Hydrates</h3>

Hydrates are compounds that contain water of crystallization with a definite mass.
These compounds are in the form of crystals and can be heated to remove water and form an anhydrous form of the compound.
Unknown hydrates with an "n" before H2O. For example; Sr(NO3)2•nH2O.
"n" represents the hydration number or the moles of water of crystalization in hydrates and can be determined using some steps.

<h3>Steps in determining hydration number in hydrates;</h3>

Determining the mass of water produced when hydrates are heated
Converting the mass of water to moles
Convert the mass of anhydrate that is left over to moles.
Find the mole ratio of anhydrate to water
Use the mole ratio to write the formula

Therefore:

We can determine the hydration in Sr(NO3)2•nH2O

<h3>Step 1: Mass of water </h3>

Mass of water = Mass of hydrate - mass of anhydrate

= 30.7 g - 22.9 g

= 7.8 g

<h3>Step 2: Moles of water </h3>

Number of moles = mass/ molar mass

= 7.8 g / 18 g/mole

= 0.433 moles

<h3>Step 3; Moles of anhydrate </h3>

Mass of anhydrate = 22.9 g

Number of moles = mass/molar mass

Molar mass of anhydrate Sr(NO3)2 is 211.63 g/mol

Thus;

Number of moles of Sr(NO3)2 = 22.9 g /211.63 g/mol

= 0.108 moles

<h3>Step 4: Mole ratio of anhydrate to water </h3>

Moles of water = 0.433 moles

Moles of anhydrate = 0.108 moles

Mole ratio = 0.108 : 0.433

= 1 : 4

Therefore; The value of n is 4

Hence the formula of the hydrate is Sr(NO3)2• 4H2O

Keywords: Hydrates, water of crystallization, steps in finding the formula of hydrates.

<h3>Learn more about: </h3>

Hydrates : brainly.com/question/10665473
Formula of hydrates example: brainly.com/question/10665473
Empirical formula: brainly.com/question/12061700
Example on empirical formula: brainly.com/question/12061700
Molecular formula: brainly.com/question/12061700

Level : High school

Subject: Chemistry

Topic: Moles

Sub-topic: Formula of hydrates

Paul [167]3 years ago

5 0

The difference of the sample mass and the heated sample mass is the amount of water which evaporated by heating process, then we can calculate:

mass of H₂O = 30.7 - 22.9 = 7.8 g

then calculate the number of moles:

number of moles of H₂O = mass / molecular mass

number of moles of H₂O = 7.8 / 18 = 0.43 moles

So the hydration number is 0.43.

Sr(NO₃)₂ • 0.43 H₂O

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maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

<u><em>In our problem:</em></u>

P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

<u><em>In our problem:</em></u>

P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0