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3 years ago

Match these solutions with their examples:

Chemistry

1 answer:

alina1380 [7]3 years ago

6 0

Answer:

sea water

Explanation:

a sea water can match the following

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How does the number of products in a synthesis reaction compare with the number of reactants?

bearhunter [10]

Its a)the no. of products are greater than the no. of reactants

5 0

4 years ago

Read 2 more answers

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

Write the balanced equation for the combustion of isooctane to produce water vapor and carbon dioxide gas.Assuming gasoline is i

tangare [24]

Answer:

The yield is 16 moles of CO2 per 2 moles of isooctane.

The mass of CO2 is 1.15 billion tonnes

Explanation:

First we set the unbalanced equation:

$C_{8}H_{18} + O_{2} \longrightarrow CO_{2} + H_{2}O$

Now we proceed to balance the equation:

$2C_{8}H_{18} + 25O_{2} \longrightarrow 16CO_{2} + 18H_{2}O$

With the equation balanced, we see that the theoretical yield is 16 moles of CO2 per 2 moles of isooctane and we can proceed to make calculations with the following data:

$\rho _{C_{8}H_{18}}=0.69 g/mL\\Consumption: 142.86x10^{9} gal\\1gal=3785.41 mL\\M_{C_{8}H_{18}}=114.23g/mole\\M_{CO_{2}}=44.01g/mole\\$

Knowing the different relations between mass, density and molar mass, we have the following results:

$V_{C_{8}H_{18}}=5.41x10^{14}mL\\m_{C_{8}H_{18}}=3.73x10^{14}g\\n_{C_{8}H_{18}}=3.27x10^{12}mole\\n_{CO_{2}}=2.62x10^{13}mole\\m_{CO_{2}}=1.15x1015g\\m_{CO_{2}}=1.15x10^{9}tonnes$