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Nadya [2.5K]
3 years ago
9

2. Calculate the standard emf of a cell that uses the Mg/Mg+2 and Cu/Cu+2 half-cell reactions at 25ºC. Write the equation for th

e cell reaction that occurs under standard-state conditions. (10 points)
Chemistry
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

2.71 V

Explanation:

Usually in the cell notation, the left side shows oxidation. So,  

Oxidation half reaction:

Mg_{(s)}\rightarrow Mg^{2+}_{(aq)}+2e^-    E^o_{ox}=-E^o_{red}=-(-2.37\ V)=2.37\ V

Reduction half reaction:

Cu^{2+}_{(aq)}+2e^-\rightarrow Cu_{(s)}    E^o_{red}=0.34\ V

Overall reaction:

Mg_{(s)}+Cu^{2+}_{(aq)}\rightarrow Mg^{2+}_{(aq)}+Cu_{(s)}

E^o_{cell}=E^o_{ox}+E^o_{red}= (2.37+0.34)\ V=2.71\ V

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2 ways to do this
a. find %Cl in CaCl2
2 x 35.45g/mole = 70.9g Cl
70.9g Cl / 110.9g/mole CaCl2 = 63.93% Cl in CaCl2
0.6963 x 145g = 92.7g = mass Cl

b. determine moles CaCl2 present then mass Cl
145g / 110.9g/mole = 1.31moles CaCl2 present
2moles Cl / 1mole CaCl2 x 1.31moles = 2.62moles Cl
2.62moles Cl x 35.45g/mole = 92.7g Cl
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3 years ago
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5. If 1 g of a gas occupies a volume of 300 mL at STP, what is the molecular weight of the gas?
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Explanation:

At STP 1 mole of an ideal gas has volume of 22,4 L. Since we know the volume of the gas we can find the number of moles of the gas. (300 mL=0,3 L)

n=0,3L/22,4 L=0,01339 mol

Since we know weight of the gas as 1 g, we can find the molecular weight as;

MW=1 g/0,01339 mol =74,67 gr/mol

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A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)

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