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nlexa [21]
4 years ago
11

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing tin

and silver: sn(s)|sn2 (aq)||ag (aq)|ag(s)
Chemistry
1 answer:
mojhsa [17]4 years ago
8 0
  The  half  reactions    as  they  occur   at  each  electrode

is  as  follows

at  the  anode  Sn(s)  =sn^2+(aq)  +  2e  -
at  the  cathode  2  ag^+(aq)   + 2e -  =  2Ag (s)

net  cell  reaction  =   Sn (s)  +  2Ag^+(aq)  =  sn^2+ (aq) +  2  Ag  (s)
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How many molecules of N204 are in 85.0 g of N2O4?
alukav5142 [94]

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

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Kruka [31]
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4 years ago
What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 11.630 g to 4.000 g ?
Dima020 [189]

The amount of sugar is 2621 mg

Why?

The complete question is:

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.

Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?

To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:

\frac{8.315 gSugar}{12.630 g Chocolate}*\frac{1000mg}{1g}=658.35mgSugar/gChocolate

Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:

4.000gChocolate*\frac{658.35mgSugar}{1gChocolate}=2621mgSugar

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