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Oksana_A [137]
2 years ago
13

Consider each possible structure of carbon dioxide, c o 2. Determine whether the structure is correct and justify your decision.

Chemistry
1 answer:
Mrrafil [7]2 years ago
6 0

This structure is correct due to the total number of bonds and electrons

We employ the octet rule, which states that all atoms in a compound are expected to follow, to check the accuracy of any chemical structure. The octet rule is precisely satisfied for each atom in the depicted structure of carbon IV oxide. The valence shell of each atom in the molecule is surrounded by eight electrons. We may thus infer that this structure is accurate given the total number of bonds and electrons since CO2 has sixteen valence electrons.

It possesses a negative charge of 1.602176634 coulombs, the basic unit of electric charge, which is comparable to its negative charge. The rest mass of an electron is 9.1093837015 10^-31 kg, or just 1/1,836 the mass of a proton. An electron is therefore considered to have practically negligible mass in comparison to a proton or neutron, and its matter is not taken into consideration when calculating an atom's mass number.

The question is incomplete, the complete question is found in the image attached to this answer

To know more about   bonds visit : brainly.com/question/13190131

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Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(
yaroslaw [1]

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

2H^+ + 2e^- \to H_2   \ \ \ \mathbf{E^0_{red}= 0.00 \ V}

At the anode:

At oxidation reaction is:

Zn \to Zn^{2+} +2e^-   \ \ \ \mathbf{E^0_{ox} = 0.76 \ V}

The overall equation for the reaction is:

\mathbf{Zn + 2H^+ \to Zn^{2+} + H_2}

The overall cell potential is:

\mathbf{E^0_{cell}= E^0_{ox} + E^0_{red}}

\mathbf{E^0_{cell}= 0.76 \ V +0.00 \ V}

\mathbf{E^0_{cell}= 0.76\ V}

Using the formula for the Nernst equation:

E = E^0 - ( \dfrac{0.0591}{n})log (Q)\\

where;

E = 0.66

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Then

0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )

0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log  [H⁺]

4.118  = - 2  log  [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

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