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3 years ago

What is the reduction half-reaction for the following unbalanced redox equation?

2 answers:

5 0

First find the oxidation states of the various atoms:
in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 
Note that N gained electrons, ie, was reduced; Cr was oxidized 
Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out 
B.NH4+ → N2 

Which of the following is an oxidation half-reaction? 
A.Sn 2+ →Sn 4+ + 2e- 
Sn lost electrons so it got oxidized

Reptile [31]3 years ago

4 0

C. N2 → NH4+ would be the correct option

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Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you

Leni [432]

Answer:

a) $\Delta G=2.6kJ$

b) $\Delta G=-979.57kJ$

c) $\Delta G=264.21kJ$

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

$\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol$

So we proceed as follows:

$\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ$

$\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ$

$\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ$

Regards.

6 0

2 years ago

In an ionic bond, electrons are ___________________. Choose all the apply.

Rainbow [258]

In an ionic bond, electrons are transferred from one stone to another atom (shared).

4 0

2 years ago

Read 2 more answers

Calculate the pressure (in kpa) of 1.5 mole of helium gas at 354 k when it occupies a volume of 16.5l.

3241004551 [841]

Answer:

267.57 kPa

Explanation:

Ideal gas law:

PV = n RT R = 8.314462 L-kPa/K-mol

P (16.5) = 1.5 (8.314462)(354) P = 267.57 kPa

8 0

1 year ago

Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0

2 years ago

Can someone plz help me with this plz?????

Otrada [13]

Answer:

We can't see the options so we don't know what we can put

Explanation:

4 0

2 years ago