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Naya [18.7K]
2 years ago
15

What type of carbon cycling occurs when stored carbon in organisms is released when they die, and when carbon from dead organism

s can be released underground in the form of coal, oil, or natural gas?
1. long-term carbon cycling
2. carbon release cycling
3. carbon storage cycling
4. short-term carbon cycling
Chemistry
1 answer:
Keith_Richards [23]2 years ago
3 0
Carbon storage cycling is the answer
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How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so
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Answer:

V H2O = 170.270 mL

Explanation:

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⇒ Q = m<em>C</em>ΔT

∴ m: mass (g)

∴ <em>C</em>:<em> </em>specific heat

assuming:

  • δ H2O = δ Coffe = 1.00 g/mL
  • <em>C</em> H2O = <em>C</em> coffe = 4.186 J/°C.g....from literature

⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)

∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe

⇒ Q  = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J

⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)

⇒ (26371.8 J)/(154.882 J/g) = m H2O

⇒ m H2O = 170.270 g

⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL

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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
ratelena [41]

Answer:

\Delta G^0 _{rxn} = 207.6\ kJ/mol

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

\Delta G_f^0  \ \ \ N_2O_{(g)} = 103 .8  \ kJ/mol

\Delta G_f^0  \ \ \ N_2{(g)} =0 \ kJ/mol

\Delta G_f^0  \ \ \ O_2{(g)} =0 \ kJ/mol

\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

K_p = 0.045

\Delta G = \Delta G^0 _{rxn} + RT \ lnK

where;

R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

7 0
4 years ago
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