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KATRIN_1 [288]
3 years ago
15

How many moles are there in 8.50 X 1024 molecules of sodium sulfate, Na2SO3?

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0
Hello!

To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:

8,50* 10^{24}molecules* \frac{1 mol Na_2SO_3}{6,02* 10^{23}molecules}=14,12 molesNa_2SO_3

So, 8,50 * 10²⁴ molecules of Na₂SO₃ represent 14,12 moles of Na₂SO₃ 

Have a nice day!
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if .40 L if water is added to the volume of cup 3, what would be the new molarity of a 2 M solution of kool-aid
stiks02 [169]

Answer:

0.7692 M ≅ 0.77 M.

Explanation:

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<em>(MV) before dilution = (MV) after dilution.</em>

M before dilution = 2.0 M, V before dilution = 0.25 L.

M after dilution = ??? M, V after dilution = 0.25 L + 0.40 L = 0.65 L.

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8 0
3 years ago
Consider the following half-reactions and their standard reduction potential values to answer the following questions.
Ganezh [65]

Answer:

Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode

Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode

Explanation:

We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.

The overall reaction equation is;

2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)

E°cell= E°cathode - E°anode

E°cathode= 1.08 V

E°anode= 0.15V

E°cell = 1.08-0.15 = 0.93 V

But

∆G°= -nFE°cell

n= 2, F=96500C, E°cell= 0.93V

∆G° = -(2× 96500× 0.93)

∆G= -179490 J

But;

∆G = -RTlnK

R=8.314 JK-1

T= 25+273= 298K

Kc= the unknown

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Substituting values and making lnK the subject of the formula

lnK= ∆G/-RT

lnK= -( -179490/8.314 × 298)

lnK= 72.45

K= e^72.45

K= 2.91×10^31

4 0
3 years ago
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