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marta [7]
3 years ago
10

Please help me with these questions

Chemistry
1 answer:
zaharov [31]3 years ago
5 0
I’m just answering this so i can ask more questions and it has to be 20 words long so i hope you figure your problems out and merry christmas happy new year
You might be interested in
Give the % composition for each element found in tin (IV) nitride.
drek231 [11]

32.3699% Tin (Sn)

15.2774% Nitrogen (N)

52.3527% Oxygen (O)

4 0
2 years ago
Plz help it is due rn, help ASAP
horrorfan [7]

Answer:

A

Explanation:

8 0
2 years ago
Read 2 more answers
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).

4 0
3 years ago
How many ammonium ions and how many sulfate ions are present in an 0.250 mol sample of (nh42so4?
Nastasia [14]
Thw answer is PHj78 JJ CP30 R2D2
7 0
3 years ago
Ammonia, NH3; ammonium nitrate, NH4NO3; and ammonium hydrogen phosphate, (NH4)2HPO4, are all common fertilizers. Rank the compou
ValentinkaMS [17]

Answer:

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

Explanation:

Mass percentage -

Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.

i.e.

mass % A = mass of A / mass of solution * 100

Given,  

mass of nitrogen = 14 g/mol

mass of hydrogen = 1 g/mol

mass of oxygen = 16 g/mol

mass of phosphorus = 31 g/mol

1. Ammonia, NH₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 1 * 14 g/mol  + 3 * 1 g/mol  

mass of solution = 17 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 17 g/mol * 100

mass % N = 82.35 % .

2.   ammonium nitrate, NH₄NO₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 4 * 1 g/mol  + 3 * 16 g/mol

mass of solution = 80 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 80 g/mol * 100

mass % N = 17.50 % .

3.   ammonium hydrogen phosphate, (NH₄)₂HPO₄

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 9 * 1 g/mol  + 4 * 16 g/mol + 1 * 31 g/mol

mass of solution = 132 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 132 g/mol * 100

mass % N = 10.60 % .

Hence , the correct order is -

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

3 0
3 years ago
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