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klio [65]
3 years ago
9

In the reaction: NH4+ + H2O ⮀ NH3 + H3O+ water is acting as a(n)

Chemistry
2 answers:
Rasek [7]3 years ago
8 0

Answer:

Water acts as a Brønsted-Lowry Base.

Explanation:

Hello,

In this case, for the theories describing acids, we say that a Brønsted-Lowry Acid is such substance which is able to donate a H⁺ and a Brønsted-Lowry base is such substance which is able to accept a donated H⁺.

In such a way, since hydronium (H₃O⁺) is a product, meaning that water received one H⁺, we conclude water is acting as a Brønsted-Lowry base. Moreover, ammonium ion acts as Brønsted-Lowry Acid.

Best regards.

Anastasy [175]3 years ago
4 0

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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d

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In a redox reaction, reduction is defined by the:
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Answer:

C. Gain in electron(s) resulting in a decrease of oxidation number.

Explanation:

Redox reactions are reactions involving transfer of of electron between two species (reduction specie) and (oxidation species) and change resulting in change in oxidation number.

Reduction in terms of redox reaction is the specie that accepts electron(s) and gets "reduced" since its oxidation state has been reduced.

For example

Cl + e- → Cl⁻

The above reaction is an example of reduction reaction taking place in a redox reaction. We can see that Chlorine oxidation state was changed from (0) to (-1) state.

5 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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