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Lynna [10]
3 years ago
9

State the name of the ion which is oxidised in the following half equations. Cathode: Na+ + e– → Na Anode: 2Cl– → Cl2 + 2e–

Chemistry
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

hahahahhahhhahahaha

Explanation:

haahahahahhahahhaha

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What does creativity have to do with Science?​
Lady_Fox [76]

creativity is the very element in science that helps us understand things. if we didn't have creative minds, we would never conduct experiments that help us discover things.

hope this helps, good luck!:)

5 0
2 years ago
If you observe with the microscope that a cell contains chloroplasts can you conclude that this is a plant cell?
shusha [124]
Obviously since plant cell contains chloroplasts.
8 0
3 years ago
At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil
sertanlavr [38]

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

6 0
3 years ago
In the following neutralization reaction, which substance is the acid?
denis-greek [22]

Hydrochloric acid is a colorless, highly pungent solution of hydrogen chloride in water. It is a corrosive

5 0
3 years ago
Read 2 more answers
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons
elena-s [515]

Answer:

-372000 J or -372 KJ

Explanation:

We have the electrochemical reaction as;

Mg(s)  +  Fe^2+(aq)→  Mg^2+(aq)  +   Fe(s)

We must first calculate the E∘cell from;

E∘cathode -  E∘anode

E∘cathode = -0.44 V

E∘anode = -2.37 V

Hence;

E∘cell = -0.44 V -(-2.37 V)

E∘cell = 1.93 V

n= 2 since two electrons were transferred

F=96,500C/(mol e−)

ΔG∘=−nFE∘

ΔG∘= -( 2 * 96,500 * 1.93)

ΔG∘= -372000 J or -372 KJ

4 0
2 years ago
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