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3 years ago

I have no idea how to do this, any help would be appreciated

Chemistry

1 answer:

aksik [14]3 years ago

6 0

Proton number = the atomic number (which is the smaller number
neutron number = the mass number (the bigger number) - the atomic number
number of electrons = the atomic number - the charge (it depends on the element but group 1 is +1 group 2 is +2 group 3 is +3 group five is -3 group six is -2 group seven is -1

i would solve the whole thing but its unclear hope this helps tho

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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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Answer:

The correct answer is 5 moles of CO2 are produced.

Explanation:

The given reaction:

CH₄ (g) + 2O₂ (g) ⇔ CO₂ (g) + 2H₂O (g)

The given reaction is an illustration of a combustion reaction. Any reaction in which a substance is burnt in excess of oxygen to generate water and carbon dioxide is termed as a combustion reaction. From the given equation, it is clear that the moles of the formation of the products are in the ratio 1: 2, that is, if 10 moles of H₂O is produced, the production of 5 moles of CO₂ will be produced.

Let us multiply, the given equation with 5 we get,

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Hence, it is clear that with the formation of 10 moles of H₂O, formation of 5 moles of CO₂ will also take place.

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