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Tcecarenko [31]
1 year ago
7

How many moles are in 8.30x10^23 molecules of H2o

Chemistry
2 answers:
sasho [114]1 year ago
8 0

Answer: 1.38 moles

Explanation: According to Avogadro's law, 1 mole of every substance contains Avogadro number of particles (atoms, molecules or ions) i.e. 6.023\times 10^{23} particles.

Given : 1 mole molecule of water contains 6.023\times 10^{23} molecules of water.

6.023\times 10^{23} molecules are present in = 1 mole

8.30\times 10^{23}  molecules will be present in =\frac{1}{6.023\times 10^{23}}\times 8.30\times 10^{23}=1.38 moles

Thus 8.30\times 10^{23} molecules is equal to 1.38 moles.

LUCKY_DIMON [66]1 year ago
4 0
You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25

That should be the right answer but I'm not sure. It has been awhile since I have done this.

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Place each charge form of alanine under the pH condition where it would be the predominant form. The pKa values for the carboxyl
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Answer:

(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH  

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(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-

(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH

Explanation:

Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3  and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:

1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).

2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).

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Explanation:

The given data is as follows.

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So,     No. of moles = Molarity × Volume

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              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

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Hence,     mass of HCl = density × Volume of HCl

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                     0.0773 = T_{f} - 18.24

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Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

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