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N76 [4]
3 years ago
6

The density of the hydrocarbon in part (a) is 2.0 g l-1 at 50°c and 0.948atm. (i) calculate the molar mass of the hydrocarbon. (

ii) determine the molecular formula of the hydrocarbon.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

1. Answer;

=56 g/mol

Explanation and solution;

PV = nRT

nRT= mass/molar mass (RT)

molar mass = (mass/V ) × (RT/P)

                   = Density × (RT/P)

Molar mass = 2.0 g/L × (0.0821 × 323 K)/0.948 atm

Molar mass = 56 g/mol


2. Answer;

Molecular mass is C4H8

Explanation;

Empirical mass × n = molar mass

Empirical mass for CH2 = 14 g/mol

Therefore;

56 g/mol = 14 g/mol × n

   n = 4

The molecular formula= 4(CH2)

    = C4H8


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2 AIPO4 + 3 BaCl2 --> 2 AICIz + 1 Baz(PO4)2
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Answer:

12.3 moles of AlCl_{3} are produced by reacting 18.4 moles of BaCl_{2} .

Explanation:

Balanced reaction: 2AlPO_{4}+3BaCl_{2}\rightarrow 2AlCl_{3}+Ba_{3}(PO_{4})_{2}

According to balanced equation:

3 molecules of BaCl_{2} produce 2 molecules of AlCl_{3}

So 3N_{A} molecules of BaCl_{2} produce 2N_{A}  molecules of AlCl_{3} (N_{A} is avogadro number).

We know 1 mol of molecule = N_{A} number of molecules.

So 3 moles of BaCl_{2} produce 2 moles of AlCl_{3}.

Hence 18.4 moles of BaCl_{2} produce (\frac{2}{3}\times 18.4) moles of AlCl_{3} or 12.3 moles of AlCl_{3} .

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The group in an experiment that is not exposed to the tested variable is called the group
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yes your answer is correct for this question.

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9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
2 years ago
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