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labwork [276]
3 years ago
13

Consider the following balanced equation: Cu + 2AgNO3 Cu(NO3)2 + 2Ag How many moles of AgNO3 would be needed to produce 2.8 mole

s of Cu(NO3)2?
Chemistry
2 answers:
S_A_V [24]3 years ago
8 0
2.8(2/1)=5.6 moles of AgNO3 would be  needed
sveticcg [70]3 years ago
4 0

Answer:

5.6 moles of AgNO3 is required to produce 2.8 moles of Cu(NO3)2

Explanation:

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

Before solving this one must make sure the equation is balanced.  From the balance equation you could notice that the AgNo3(silver nitrate) has 2 moles while the Cu(NO3)2 has 1 mole.

If 2 moles = 1 mole

    ?   = 2.8 moles

cross multiply 2.8 x 2 = 5.6 moles    

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If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
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Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

c'=2.83\times 10^{-6} M

The final concentration of the diluted solution is 2.83\times 10^{-6} M

8 0
3 years ago
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