An exothermic reaction is what energy.

Describe the makeup of Earth’s atmosphere. How do solids get into the air?

Alex17521 [72]

Earth's atmosphere is 78% nitrogen, 21% oxygen, 0.9% argon, and 0.03% carbon dioxide with very small percentages of other elements. Our atmosphere also contains water vapor. In addition, Earth's atmosphere contains traces of dust particles, pollen, plant grains and other solid particles.

5 0

4 years ago

Read 2 more answers

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

5.

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago

Which planet formed near the sub where the solar system l’s temperatures were very high

Harrizon [31]

Answer:

Mars

Explanation:

Terrestrial or inner planets like Mars and Venus were formed near the Sun where the solar system's temperatures were very high.

4 0

3 years ago

Read 2 more answers

You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work

Bogdan [553]

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n * R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm

Hope that helps!

7 0

4 years ago

A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

5 0

4 years ago