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grandymaker [24]
3 years ago
6

What is the percent composition of chromium in CrCl3?

Chemistry
1 answer:
maks197457 [2]3 years ago
8 0

The oxidation state of Chromium chloride (III) is +3

How can we find the answer?

First of all write down what you know about the molecule:

1: The molecule hasn't got an electric charge (is not an ione), this means that either positive and negative charges of its atoms are balanced (we have the same number of positve and negative charges)

2: Since it's a salt, where the metal is chromium and the non metal is the alogen Chlorine, we know that the negative charge belogns to the non metal element because of its elettronegativity, therefore the positive charge belongs to the metale element (chromium).

3: when chlorin forms binary salts its oxidation state is always -1 (you can find out this info in a periodic table)

In <span>CrC<span>l3</span></span> we have 3 chlorine atoms where each of them carrys 1 negative charge, so the total amount of negative charges is -3

Since the charges are balanced, the question is: Which is the positive charge that Chromium must carry in order to balance 3 negative charges?

The answer comes out to +3


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What characterizes a Homo genesis mixture
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Answer:

The correct answer would be the third choice.

Homo-genesis mixtures are the same in composure, and will be hard to break apart other then with chemical means.

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6 0
3 years ago
A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

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