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Luda [366]
2 years ago
6

What percentage of oxygen is attributed fossil fuel combustion

Chemistry
1 answer:
Oxana [17]2 years ago
6 0

Answer:

C. 4%

Explanation:

The answer is 4%

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ANSWER- ocean winds can carry moisture with then mmm and can bring rain.

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How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of
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14.7 lbs

Explanation:

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3 years ago
What is the result of the following calculation? Record to the correct number of significant figures.
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8.15103

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3 0
2 years ago
What is the pH of 2.5×10 to the power of negative2IM solution of HCL?
Archy [21]

Answer:

pH=1.6

Explanation:

Recall that pH= -log [H+]

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Hoped this helped!

3 0
3 years ago
5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas
Artist 52 [7]

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)

where,

\Delta U = change in internal energy = ?

n = number of moles of gas = 5 moles

C_v = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta U=nC_v(T_2-T_1)

\Delta U=(5moles)\times (2R)\times (573-298)

\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)

\Delta U=22863.5J=22.86kJ

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)

where,

\Delta H = change in enthalpy = ?

n = number of moles of gas = 5 moles

C_p = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta H=nC_p(T_2-T_1)

\Delta H=(5moles)\times (3R)\times (573-298)

\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)

\Delta H=34295.25J=34.29kJ

The change in enthalpy of the gas is 34.29 kJ.

4 0
3 years ago
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