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Scorpion4ik [409]
2 years ago
7

How many molecules of CBr4 are in 250 grams of CBr4

Chemistry
1 answer:
Kazeer [188]2 years ago
8 0

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


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Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

                             24 \times 3600 sec

Hence, total charge passed daily is calculated as follows.

                      150,000 \times 24 \times 3600 sec

And, number of Faraday of charge is as follows.

                    \frac{150,000 \times 24 \times 3600 sec}{96500}

                     = 134300.52 F

The oxidation state of aluminium in Al_{2}O_{3} is +3.

                       Al^{3+} + 3e^{-} \rightarrow Al(s)

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

                \frac{134300.52 F}{3} \times \frac{89}{100}

                   = 39842.487 mol

or,               = 3.9842 \times 10^{4} mol

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

            Mass in grams = 3.9842 \times 10^{4} mol \times 27

                                     = 107.57 \times 10^{4} g

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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
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qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

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The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

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Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

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Finally, solve for Tfinal:

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The answer should have three significant figures, so round to 22.7∘C.

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