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brilliants [131]
3 years ago
14

What is the empirical formula of a compound containing 40.0% sulfur and 60.0% oxygen by mass?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

in order to determine empirical formula we have to determine the mole ratio of the given elements

Let the total mass of the compound is 100g

as given that the compound has 40% sulfur , so mass of sulfur = 40g

as given that the compound has 60% oxygen, so mass of oxygen = 60g

let us calculate the moles of each element

Moles of sulfur = mass / atomic mass = 40 / 32 = 1.25

moles of oxygen = mass / atomic mass = 60/ 16 = 3.75

In order to get simple ratio of moles we will divide both the moles with least number of moles which is 1.25

moles of sulfur = 1.25 / 1.25 = 1

moles of oxygen = 3.75 /1.25 = 3

So empirical formula will be SO₃

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4 0
2 years ago
The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

  • XSO₂ = 0.58/3.29 = 0.176
  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

5 0
3 years ago
Definition of matter​
Contact [7]

Answer:

Any thing that occupied volume and mass in air is known as matter

4 0
3 years ago
Read 2 more answers
Someone please help! this is the last question<br>I only need help with B.<br><br>​
ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

6 0
2 years ago
A solution has a oh- = 1 x 10-5 m what are the h30 and the ph of the solution
Deffense [45]

I think it's easiest to find the pOH from the given [OH-] first.

-log(1x10^-5)

pOH=5

Then find the pH.

pOH+pH=14

5+pH=14

pH=9

Then find the [H+] using the pH.

antilog(-9) (if you dont have an antilog button use 10^-9)

[H+]=1x10^-9

3 0
3 years ago
Read 2 more answers
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