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V125BC [204]
3 years ago
12

suppose work input is 25 j and the output distance is 10 m factoring in the effect of friction which must be true about output f

orce a) it equals 2.5n b) it is less than 2.5n c) it is greater than 2.5n
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

Therefore F ∠ 2.5N

Explanation:

Definition of Work:

It is the amount of energy transferred.

Mathematical expression:

W = F × d

By taking the friction into count it will define as:

Input work = work against friction ×  output work

No we will put the values:

25 J = w + 10 m

j = kg m²/s²

25 kg m²/s² =  w + 10 m

when friction is exist then w > 0

25 kg m²/s² - 10 m  > 0

15 N > 0

Therefore F ∠ 2.5N

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1.4 mols

4th answer

Explanation:

22. 5 g of O2 in moles = (22.5/32) mols = 0.703 mol

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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ
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<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules

<u>Explanation:</u>

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E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

NH_4 will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

<u>Oxidation half reaction:</u>  NH_4\rightarrow NO_2^-+6e^-     ( × 4)

<u>Reduction half reaction:</u>  O_2+4e^-\rightarrow 2H_2O   ( × 6)

<u>Overall reaction:</u> 4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O

We are given:

Molecules of NH_4 = 28

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Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

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