What is the result at the end of meiosis II?

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det

jeyben [28]

Answer:

0.054 mol O

Explanation:

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample?

Step 1: Given data

Chemical formula of acetic acid: CH₃CO₂H

Moles of carbon in the sample: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula, the molar ratio of C to O is 2:2.

Step 3: Calculate the moles of oxygen in the sample

We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.

0.054 mol C × (2 mol O/2 mol C) = 0.054 mol O

6 0

2 years ago

What are three ways of the atmosphere is important to living things?

UNO [17]

1. Blocks Ultra violet rays that can be harmful.
2. Provides Oxygen, so we can live.
3. It blocks meteors from hitting Earth's surface.

7 0

2 years ago

HELP HURRRRYYY

LekaFEV [45]

I would say the answer is... C. AgNO3 + LiOH AgOH + LiNO3

Good luck!!

5 0

2 years ago

PLEASE HELP ME ASAP!

uranmaximum [27]

your answer is b hope this helps

7 0

3 years ago

Read 2 more answers

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