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Natali [406]
2 years ago
13

If the cell is somehow operated under conditions in which it produces a constant voltage of 1.50 V , how much electrical work wi

ll have been done when 0.487 mL of Br2(l) has been consumed
Chemistry
1 answer:
Yuri [45]2 years ago
8 0

For a cell  operated to produce a constant voltage of 1.50 V, the electrical work done  is mathematically given as

W=2.077*10^{-7}

<h3>What is the electrical work done?</h3>

Generally, the volume of Br2  is mathematically given as

Vb=\frac{0.38*3.12}{159.809}

Vb=7.18*10^-3ml

Therefore, the quantity of charge

q=(1.436*10^-2)*9.644*10^4

q=1.385^{-4}c

In conclusion, work done

W=Eq

W=1.50*1.385^{-4}

W=2.077*10^{-7}

Read more about Work

brainly.com/question/756198

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Answer:

c. tertiary.

Explanation:

In this case, we can review the definition of each level of structuration in the proteins:

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In the primary structure, the amino acids are linked by peptide bonds. That is, the order of the amino acids is the criterion that defines this type of structure.

<u>Secondary structure</u>

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In the secondary structure, we have to look at the way in which the protein is folded. The options are:

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<u>Tertiary structure</u>

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In the tertiary structure, the R groups that the amino acids have in the primary structure can generate interactions with each other. Interactions such as hydrogen bridges, dipole-dipole, hydrophobic interactions. This makes the protein have a very specific three-dimensional structure, on which its function depends.

<u>Quaternary structure</u>

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In the quaternary structure, several subunits may be attached, or there may be prostatic groups (metals that can help to attach various protein units).

With all these in mind, the deffinition that fits with the description in the question is the <u>tertiary structure.</u>

I hope it helps!

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What is the number of molecules of c2h5oh in a 3m solution that contains 4.00kg h2o?
Veronika [31]
 The number  of C2H5OH  in a 3 m solution that contain 4.00kg H2O is calculate as below

M = moles of the solute/Kg  of water

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=12 moles/1 moles  x 6.02 x10^23 = 7.224 x10^24 molecules
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