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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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3 years ago
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grandymaker [24]

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3 0
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Read 2 more answers
Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B
eimsori [14]
2 boxes of A
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3 0
3 years ago
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0
2 years ago
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