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muminat
3 years ago
13

As a cooking method, braising is valued for its A. ability to retain flavor. B. quickness and convenience. C. ability to create

delicate flavors. D. simplicity and versatility.
Chemistry
1 answer:
Setler [38]3 years ago
5 0
Definitely C, ability to create delicate flavors. :)
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Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some
Gekata [30.6K]

Answer:

k_{sp}=4.7 \times 10^{-13}.

Explanation:

Now equation of tqo halves are:

Oxidation : Ag(s)-->Ag^+(aq)+e^-

Reduction : AgBr(s)+e^--->Ag(s)+Br^-(aq)

We know,

E^o_{cell}=0.071-(0.8)=-0.729\ V.

\Delta G^o=-n\times F \times E^o= -R\times T\times ln(k_s_p)\\-1\times 96485\times (-0.729)=-8.314\times 298\times ln(k_s_p)\\k_s_p=4.7\times 10^{-13}.

k_{sp}=4.7 \times 10^{-13}.

Hence, this is the required solution.

5 0
3 years ago
Octane, c8h18, is a component of fuel used in internal combustion engines. the dominant intermolecular forces in octane are
lesya692 [45]
<span>London dispersion forces</span>
6 0
3 years ago
NaCl + H2SO4 ---&gt; Na2SO4 + HCl<br> Balance the double replacement chemical reaction.
vlada-n [284]

Answer:

2NaCl + H2SO4 ----> Na2SO4 + 2HCl

Explanation:

4 0
3 years ago
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Which real world scensories below respent physical changes check all that apply
Delicious77 [7]

Answer:

I need more information

Explanation:

4 0
3 years ago
What is the cell potential of an electrochemical cell that has the half -reactions shown below? Ni^ 2+ +2e Ni A| A|^ 3+ +3e^ -
Nadya [2.5K]
<h3>Answer:</h3>

A. 1.4 V

<h3>Explanation:</h3>

We are given the half reactions;

Ni²⁺(aq) + 2e → Ni(s)

Al(s) → Al³⁺(aq) + 3e

We are required to determine the cell potential of an electrochemical cell with the above half-reactions.

E°cell = E(red) - E(ox)

From the above reaction;

Ni²⁺ underwent reduction(gain of electrons) to form Ni

Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺

The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V

Therefore;

E°cell = -0.25 V - (-1.66 V)

         = -0.250V + 1.66 V

         = + 1.41 V

         = + 1.4 V

Therefore, the cell potential will be +1.4 V

3 0
3 years ago
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