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asambeis [7]
3 years ago
10

When the free-energy change (AG) of a reaction is negative, the reaction is said to be: Endergonic Non spontaneous Spontaneous E

ndothermic
Chemistry
1 answer:
Greeley [361]3 years ago
6 0

Answer:

Spontaneous

Explanation:

For a spontaneous reaction , the value for the change in free energy or gibbs' free energy is negative .

The change in free energy , i.e. , ΔG , denotes the maximum amount of usable energy released , as going from initial state , i.e. , the reactant towards the final state , i.e. , the product .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

ΔG < 0 , the reaction is Spontaneous

ΔG > 0 , the reaction is non Spontaneous

ΔG = 0 , the reaction is at equilibrium

For ,

ΔG < 0 , that the reaction proceed without any energy input , hence , it is  Spontaneous in nature .

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Sam believes the rock is more denser than the pencil. He measured the mass of the rock to be 8.5 grams and volume to be 4.5mL. H
mezya [45]

Answer:

Rock

Explanation:

Let's calculate the density of each object:

Rock:

Density = \frac{mass}{volume}=\frac{8.5\ g}{4.5\ mL}=1.9\ g/mL

Pencil:

Density = \frac{mass}{volume}=\frac{2.4\ g}{4.5\ mL}=0.53\ g/mL

Therefore the rock is denser.

6 0
2 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

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

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
PLEASE HELP AND THANKS
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Answer:

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4 0
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What can be expected to occur as climate change continues on earth?
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3 years ago
How is kinetic energy related to mass and velocity?
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