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Bess [88]
3 years ago
14

Explain why the energy harnessed per person (per capita) has increased so dramatically since the industrial revolution.

Chemistry
2 answers:
Mama L [17]3 years ago
4 0

Answer:

Increased energy use leads to increased economic activity and an increased quality of life. Ever since the industrial revolution, advances in technology have let humans in industrialized nations increase their standard of living significantly, which required significant increases in the amount of energy harnessed per capita.

Explanation:

tell me if yall agree or not please i want to help yall.

qwelly [4]3 years ago
3 0
<span>"Increased energy use leads to increased economic activity and an increased quality of life. Ever since the industrial revolution, advances in technology have let humans in industrialized nations increase their standard of living significantly, which required significant increases in the amount of energy harnessed per capita."


</span>
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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

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4 years ago
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_Pb(OH)2+_HCl -&gt; _H2O+_PbCl2
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20 + CIPb

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