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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