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Ksenya-84 [330]
3 years ago
14

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
Chemistry
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

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Answer:

8.34

Explanation:

1) how much moles of NH₃ are in the reaction;

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all the details are in the attachment; the answer is marked with red colour.

Note1: M(NH₃) - molar mass of the NH₃, constant; M(H₂) - the molar mass of the H₂, constant; ν(NH₃) - quantity of NH₃; ν(H₂) - quantity of H₂.

Note2: the suggested solution is not the shortest one.

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3 years ago
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Answer:

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Explanation:

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2 years ago
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Sunny_sXe [5.5K]

Answer:

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺   →   Ag      

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ →  Ag   (1)

Now, we need to balance the half-reaction of Cr:

Cr   →  CrO₄²⁻  

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:          

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻  

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Cr + OH⁻ →  CrO₄²⁻ + 6e⁻ + H₂O    

The balanced equation is:

Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O   (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ →  Ag)  

<u>Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O</u>

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O                  

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

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