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VLD [36.1K]
2 years ago
15

If 0.640 g of beautiful blue crystals of azulene is dissolve in 99 g of benzen, the resulting solutions boils at 80.23 degrees c

elsius. find the molecular formula?
Chemistry
1 answer:
devlian [24]2 years ago
6 0

This problem handles<em> boiling-point elevation</em>, which means we will use the formula:

ΔT = Kb * m

Where ΔT is the difference of Temperature between boiling points of the solution and the pure solvent (Tsolution - Tsolvent). Kb is the ebullioscopic constant of the solvent (2.64 for benzene), and m is the molality of the solution.

Knowing that benzene's boiling point is 80.1°C, we <u>solve for m</u>:

Tsolution - Tsolvent = Kb * m

80.23 - 80.1 = 2.64 * m

m = 0.049 m

We use the definition of molality to <u>calculate the moles of azulene</u>:

0.049 m = Xmoles azulene / 0.099 kgBenzene

Xmoles azulene = 4.87 x10⁻³ moles azulene

We use the mass and the moles of azulene to<u> calculate its molecular weight</u>:

0.640 g / 4.875 x10⁻³ mol = 130.28 g/mol

<em>A molecular formula that would fulfill that molecular weight</em> is C₁₀H₁₀. So that's the result of solving this problem.

The actual molecular formula of azulene is C₁₀H₈.

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A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo
Svetlanka [38]

Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃  +  MgCO₃ +  4HCl  →  MgCl₂  +  2NaCl  + 2CO₂  +  2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2  = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %

3 0
1 year ago
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7 0
2 years ago
How many moles of oxygen atoms are in 132.2 g of MgSO4?
zzz [600]

4.4moles of oxygen atoms

Explanation:

Given parameters:

Mass of MgSO₄ = 132.2g

Unknown:

Number of moles of oxygen atoms = ?

Solution:

The number of moles is the quantity of substance that contains the avogadro's number of particles.

 To solve for this;

 Number of moles = \frac{mass}{molar mass}

Molar mass of MgSO₄ = 24 + 32 + 4(16) = 120g/mole

  Number of moles = \frac{132.2}{120} = 1.1 moles

In

     1 moles of MgSO₄ we have 4 moles of oxygen atoms

    1.1 moles of MgSO₄ contains 4 x 1.1 moles = 4.4moles of oxygen atoms

learn more:

number of moles  brainly.com/question/1841136

#learnwithBrainly

8 0
2 years ago
For each pair below, select the sample that contains the largest number of moles. Pair A 2.50 g O2 2.50 g N2
raketka [301]

Answer:

Explanation:

Pair  2.50g of O₂ and 2.50g of  N₂

The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:

Number of moles = \frac{mass }{molarmass}

Atomic mass of O = 16g and N = 14g

Molar mass of O₂ = 16 x 2 = 32gmol⁻¹

Molar mass of N₂ = 14 x 2 = 28gmol⁻¹

Number of moles of O₂ = \frac{2.5}{32} = 0.078mole

Number of moles of N₂ = \frac{2.5}{28} =  0.089mole

We see that N₂ has the largest number of moles

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