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sergey [27]
3 years ago
13

A hydrocarbon contains C,H, and F atoms. A 5.43 g sample of this hydrocarbon was analyzed and the mass of 2.35 g of carbon and 0

.294 g of hydrogen. The molar mass of this compound was measured between 219-225 g/mol, what is the molecular formula of the hydrocarbon that contains C,H and F?
Chemistry
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

C8H12F6

Explanation:

To solve this question we need to find the moles of each atom in order to find the empirical formula (The empirical formula is defined as the simplest whole-number ratio of atoms present in a molecule). Using the empirical formula and the molar mass we can find molecular formula as follows:

<em>Moles C:</em>

2.35g * (1mol / 12g) = 0.1958 moles C

<em>Moles H:</em>

0.294g (1mol /1g) = 0.294 moles H

<em>Moles F -Molar mass: 19.0g/mol-: </em>

Mass F: 5.43g - 2.35g C - 0.294g H = 2.786g F * (1mol / 19.0g) = 0.1466 moles F

The moles of atoms dividing in the moles of F (Lower number of moles) produce the simplest ratio as follows:

C = 0.1958mol C / 0.1466mol F = 1.33

H =0.294mol H / 0.1466mol F = 2

F = 0.1466 mol F / 0.1466mol F = 1

As the empirical formula requires whole numbers, this ratio multiplied 3 times:

C = 4

H = 6

F=3

The empirical formula is:

C4H6F3

With molar mass of:

4C = 12*4 = 48

6H = 1*6 = 6

3F = 19*3 = 57

The molar mass is: 48g/mol + 6g/mol + 57g/mol = 111g/mol

As we know the molecule has a molar mass between 219-225g/mol and the empirical formula is 111g/mol, 2 times empirical formula will produce a molecule with the molar mass of the molecule, that is:

<h3>C8H12F6</h3>
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Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

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4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

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3 years ago
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DedPeter [7]

Answer:

V=14373.33\ \text{mm}^3

Explanation:

It is given that,

The radius of the inflated balloon is 7 cm.

We need to find the volume of air inside the balloon in milliliters.

The balloon is in the shape of a sphere whose volume is given by :

V=\dfrac{4}{3}\pi r^3\\\\\text{Put r = 7 cm}\\\\V=\dfrac{4}{3}\times \dfrac{22}{7}\times 7\times 7\times 7\\\\V=1437.33\ \text{cm}^3\\\\\text{We know that, 1 cm = 10 mm}\\\\\text{So},\\\\V=14373.33\ \text{mm}^3

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