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Ghella [55]
2 years ago
5

Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00

0 and 100,000 g/mola indicated below: (A) Equal numbers of molecules of each sample (B) Equal masses of each sample. 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:
AnnZ [28]2 years ago
7 0

Answer:

Hello some parts of your question is missing

Two 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.

Answers:

a) 46.666.66 g/mol

b) 20930.23 g/mol

c)43333.33 g/mol

Explanation:

A)The equal number of molecules of each sample can be calculated using  Mn = \frac{n(M1 + M2 + M3)}{3n}

because for the number of molecules to be equal : n1 = n2 = n3 = n

Mn = 46666.66 g/mol

B ) To calculate the equal masses of each sample

we apply this equation

Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3}  }

ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION

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8 0
1 year ago
A mixture of three gases has a total pressure at 298 K of 1560 mm Hg. the mixture is analyzed and is found to contain 1.50 mol N
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Answer:

the partial pressure of Xe is 452.4 mmHg

Explanation:

Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.

Molar fraction of Xe is calculated as follows:

x_{Xe}=\frac{n_{Xe} }{n_{t} }

x_{Xe}=1.75/5.9\\x_{Xe}=0.29

Then, 0.29 is the molar fraction of Xe in the mixture of gases given.

To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:

Partial Pressure of Xe=1560mmHg*0.29

Partial Pressure of Xe=452.4mmHg

7 0
1 year ago
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
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Answer:

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Explanation:

Based in the reaction:

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2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = <em>4x10⁻⁴ mol / (Ls)</em>

Right answer is:

C. 4x10⁻⁴ mol / (Ls)

8 0
2 years ago
how many moles of potassium would you need to prepare 1200 grams of 5.0% potassium sulfate (m/m) solution
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calculation
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%M/M =  mass  of the  solute(potassium)/mass of the  solvent (potassium  sulfate solution)

let  the  mass  of  potassium be represented by  Y

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Y=  60 g

moles of potassium =mass/molar  mass

= 60/39=1.538
7 0
2 years ago
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