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irakobra [83]
2 years ago
6

Please help! I will give 100pts and brainliest!!!

Chemistry
2 answers:
kirza4 [7]2 years ago
8 0

Answer:

What do you need help with?

Explanation:

Ainat [17]2 years ago
3 0

Answer:

there is no question...please attache it...

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in acetic acid ,there are 4 hydrogen atoms

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C H 3 C O O H

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Read 2 more answers
What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?
Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

4 0
3 years ago
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