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stiv31 [10]
3 years ago
9

What is the partial pressure of oxygen gas in a mixture which contains 540 mmHg He, 203 303 Pa N2 if the total pressure of the m

ixture is 5.00 atmospheres?
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

Partial pressure O₂  = 2.284 atm

Explanation:

This is problem with unit conversion. Let's convert everything to atm

760 mmHg ___ 1 atm

540 mmHg ___ (540 / 760) = 0.710 atm

101325 Pa ____ 1 atm

203303 Pa ____ (203303 / 101325) = 2.006 atm

Total pressure of the mixture = Sum of partial pressure of each gas

0.710 atm (He) + 2.006 (N₂) + Partial Pressure O₂ = 5 atm

Partial pressure O₂ = 5 atm - 2.006 atm - 0.710 atm = 2.284 atm

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The Earth’s atmosphere is divided into five layers. In which layer does all life exist?
n200080 [17]

Answer:

1. troposphere

2. less dense/decreases

3. ozone

4. nitrogen

5. xylem

6. flagellum

7. kingdom fungi

8. Ocean convection currents

9. As they fly, insects spread pollen that sticks to their bodies from the flowers.

10. the plastic handle, because it is a good insulator.

11. The hyphae, or feeding structures, reach deep into the wood to obtain nutrients

12. chemical energy

13. created or destroyed.

14. electromagnet

15. c

16. kinetic, elastic potential

5 0
3 years ago
Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
How do plants affect the amount of carbon in Earth’s atmosphere?
kaheart [24]

Answer:

c

Explanation:

plants decrease the levels of carbon dioxide in the atmosphere due to the process of photosynthesis

6 0
2 years ago
What is the formula for manganese(III) oxide?<br> Explain.
Naddik [55]

Answer:

Mn2O3

Explanation:

Manga has a 3+ charge and oxygen has a 2- charge so to balance the charges there needs to be 3 oxygens for every 2 manga

6 0
2 years ago
50 POINTS PLEASE HELP!
Aleks04 [339]

Answer: The molar mass of the gas is 9.878 g/mol.

Explanation:

According to Graham's law, the rate of diffusion is inversely proportional to square root of molar mass of gas.

Rate = \frac{1}{\sqrt{M}}

where,

M = molar mass of gas

As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.

\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\

where,

M_{1} = molar mass of hydrogen gas

M_{2} = molar mass of another given gas

R_{1} = rate of diffusion of hydrogen

R_{2} = rate of diffusion of another given gas = \frac{1}{7}R_{1}

Substitute the values into above formula as follows.

\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\\frac{R_{1}}{\frac{1}{7}R_{1}} =  \sqrt{\frac{M_{2}}{2}}\\7 \times 1.414 = M_{2}\\M_{2} = 9.878 g/mol

Thus, we can conclude that the molar mass of the gas is 9.878 g/mol.

7 0
3 years ago
Read 2 more answers
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