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slamgirl [31]
3 years ago
11

What is a term called gradient in Earth Science?

Chemistry
1 answer:
goblinko [34]3 years ago
4 0
The gradient is the slope of a linear equation, represented in the simplest form as y = mx + b. In Earth Science, the gradient is usually used to measure how steep certain changes in elevation are.

In order to find the gradient in a topographical setting, one must know two things: the elevation of two points and the distance between the two points. Once these values are known, the gradient can be found by dividing the change in field value, or the change in elevation, by the distance. The higher the gradient value is, the steeper the slope is.

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Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are
mote1985 [20]

Answer:

Unfavorable reactions: Reaction A and Reaction B.

Coupled reactions favorable: Reactions B and C and Reactions A and C.

Net change:

Reactions B and C : -14.2kJ/mol

Reactions A and B : 30.1kJ/mol

Reactions A and C: -16.7kJ/mol

Explanation:

A reaction is thermodynamically favorable (spontaneous) if ΔG < 0. Thus, the unfavorable reactions -ΔG > 0- are:

Reaction A and reaction B.

When reaction C is coupled with reaction B and reaction A the chemical equation is:

ATP + fructose-1,6-phosphate ⟶ ADP + fructose-1,6-bisphosphate

ΔG = 16.3 - 30.5 = -14.2 kJ/mol

ATP + glucose ⟶ ADP + glucose-6-phosphate

ΔG = 13.8 - 30.5 = -16.7 kJ/mol

The coupled reaction of A and B has a change in free energy of:

ΔG = 13.8 + 16.3 = 30.1 kJ/mol

4 0
3 years ago
2. What do you think a
Monica [59]

Answer:

hey discover an

interesting pattern in the

world that no one else

has described, but they

cannot explain what is

causing the pattern?

Explanation:

3 0
2 years ago
Read 2 more answers
Which one is the least dangerous to humans?
3241004551 [841]

Answer:

A is correct

I guess

Explanation:

hope this helps

7 0
2 years ago
Water’s boiling point would be close to
aleksandrvk [35]
Waters boiling point is 100°C and 212°F
4 0
3 years ago
An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
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