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Law Incorporation [45]
3 years ago
10

Which of the following is a way carbon enters the atmosphere

Chemistry
1 answer:
zaharov [31]3 years ago
3 0
Volcanic activity, obviously by erupting volcanoes
You might be interested in
I need help ASAP
vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea
Aleksandr [31]

Answer:

ΔS° = - 47.2 J/mol.K

Explanation:

  • P4O10 + 6H2O → 4H3PO4

ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10

∴ S°mH2O(l) = 69.9 J/mol.K

∴ S°mP4O10 = 231 J/mol.K

∴ S°mH3PO4 = 150.8 J/mol.K

⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231

⇒ ΔS° = - 47.2 J/mol.K

5 0
3 years ago
Read 2 more answers
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
Consider the following intermediate reactions.
Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

4 0
3 years ago
Read 2 more answers
Please help asappp‍♀️!
solniwko [45]
Can you show me the passage?
6 0
3 years ago
Read 2 more answers
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