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lianna [129]
1 year ago
6

The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?

Chemistry
2 answers:
seropon [69]1 year ago
7 0
According to the equation of molarity:

Molarity= no.of moles / volume per liter of Solution

when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L

by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
                        = 0.58 * 0.150 = 0.087 Moles
Sergio039 [100]1 year ago
4 0
Molarity is the number of moles that are contained in 1000 cm³ or a liter of water.
In this case the molarity is 0.58 M meaning 0.58 moles of the dye are contained in 1000 cm³ or a liter of water.
The beaker is at  150 cm³, there the number of moles will be;
=0.58 × 150/1000
= 0.087 moles
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damaskus [11]

Answer:

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(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-

(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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8 0
2 years ago
Assuming ideal behavior, how many moles of argon would you need to fill a 14.0×12.0×10.0 ft room? assume atmospheric pressure of
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First let us get the volume V:

volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

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volume = 1680 ft^3 * (1 m / 3.28 ft)^3 = 47.61 m^3

 

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4 0
1 year ago
6K + B2O3 → 3K2O + 2B
atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

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