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MAVERICK [17]
3 years ago
12

If the Large Hadron Collider is 27 km long, and visible light travels 3.0 × 108 m/s, how long does it take a ray of visible ligh

t to travel from one end of the collider to the other?
Chemistry
2 answers:
marysya [2.9K]3 years ago
8 0

Answer:

C) 9.0 x 10^-5 s

Valentin [98]3 years ago
6 0
27 km = 27000m. Time = distance/speed = 27000/3x10^8 = 9x10^-5 s. 
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For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
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Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

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From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

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Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

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