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Afina-wow [57]
3 years ago
6

What are bonds in which valence electrons are shared

Chemistry
2 answers:
Anton [14]3 years ago
8 0

Answer:

covalent bonds are bonds in which valence electrons are shared.

Explanation:

Sveta_85 [38]3 years ago
7 0

Answer:true

Explanation:

<u><em>Covalent bonds are a class of chemical bonds where valence electrons are shared between two atoms, typically two nonmetals. The formation of a covalent bond allows the nonmetals to obey the octet rule and thus become more stable.</em></u>

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Enough of a monoprotic acid is dissolved in water to produce a 0.0121 M solution. The pH of the resulting solution is 6.49. Calc
Hitman42 [59]
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
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3 years ago
B. The following reaction takes place in a basic solution. (7 points)
Natali5045456 [20]

<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

  • Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

The half-reactions now become:

Oxidation half-reaction: 3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-

Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

  • Add the equations and simplify to get a balanced equation

Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

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