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Talja [164]
3 years ago
5

The overall charge of an atom is ________ if the number of electrons is ________ than the number of protons.

Chemistry
1 answer:
miss Akunina [59]3 years ago
8 0
The overall charge of an atom is positive if the number of electrons is less than the number of protons. Electrons are negatively charged particles while protons are positively charged particles. If there are less electrons as compared to the number of protons in an atom, then naturally the overall charge of the atom would be positive since there are excess protons. Another case would be that the overall charge of an atom is negative if the number of electrons is greater than the number of protons. An atom having more electrons than the number of the protons present would lead to a negative value of the charge since there are excess electrons.
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\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

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\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

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Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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