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nevsk [136]
3 years ago
15

neon has an average atomic mass of 20.2 g/mol, whereas argon has an average atomic mass of 40.0 g/mol. How would the number of a

toms in a 1.0 mol sample of neon compare to the number of atoms in a 4.0 mol sample of argon
Chemistry
2 answers:
topjm [15]3 years ago
8 0
Argon sample would have 20.2 divided by 40.0
Greeley [361]3 years ago
6 0

Answer : The number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

Explanation :

As we know that the 1 mole of substance contains 6.022\times 10^{23} number of atoms.

As per question,

The 1 mole sample of neon contains 6.022\times 10^{23} number of atoms.

And,

As, 1 mole sample of argon contains 6.022\times 10^{23} number of atoms

So, 4.0 mole sample of argon contains 4.0\times 6.022\times 10^{23}=24.088\times 10^{23} number of atoms

From this we conclude that, the number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

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A polymer sample combines five different molecular-weight fractions, each of equal weight. The molecular weights of these fracti
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Answer:

Mn = 43,783

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Explanation:

The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. This means their Mi is respectively: (The molar weight (Mi) of the fractions)

Fraction 1 : Mi = 20  *10^3

Fraction 2: Mi = 40 *10^3

Fraction 3 : Mi = 60 *10^3

Fraction 4: Mi = 80 *10^3

Fraction 5 : Mi = 100  *10^3

The ΣMi = 300*10^-3

The Wi (mass of the fractions is for all the fractions the same, let's say 1)

So Wi = 1+1+1+1+1 = 5

Since number of moles = mass / Molar mass

The number of moles is respectively: ni = Wi/Mi (x10^5)

Fraction 1 : ni = Wi/Mi = 1/20000 = 5

Fraction 2: ni = 1/40000 = 2.5

Fraction 3 : ni =1/60000 = 1.67

Fraction 4: ni = 1/80000 = 1.25

Fraction 5 : ni= 1/100000 = 1

The Σni = 11.42

Mn = ΣWi/ni = 5/11.42*10^-5 = 43,783

Mw = (ΣWi * Mi)/ΣWi  = 300,000 /5 = 60,000

Mz = (ΣWi * Mi²)/ΣWi *Mi = (4*10^8 +16*10^8 +36*10^8 +64*10^8 +100*10^8) /300,000  =73,333

Mz/Mn = narrow distribution =60,000/43,783 = 1.37

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