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3 years ago

HELP ASAP!!!!!!!!

1 answer:

4 0

Answer is: c. 80 g of Br2.
1) n(Li) = 14 g ÷ 7 g/mol
n(Li) = 2 mol.
N(Li) = 2 mol · 6·10²³ 1/mol.
N(Li) = 1,2·10²⁴.
2) n(He) = 4 g ÷ 4 g/mol.
n(He) = 1 mol.
N(He) = 1 mol · 6·10²³ 1/mol.
N(He) = 6·10²³.
3) n(Br₂) = 80 g ÷ 160 g/mol.
n(Br₂) = 0,5 mol.
N(Br₂) = 0,5 mol · 6·10²³ 1/mol.
N(Br₂) = 3·10²³.
4) n(H₂) = 4 g ÷ 2 g/mol = 2 mol.
N(H₂) = 1,2·10²⁴.

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3 years ago

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Explanation:

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If 8 atoms of Hydrogrn are present in 1 molecule of propane.

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- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

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$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

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