$1 \times 10^{21}$ $1 xx 10^21$ molecules are removed from $280 mg$ $280" mg"$ of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer: $8.4 \times 10^{- 3} moles$ $8.4 xx 10^- 3" moles"$ ]

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$1 \times 10^{21}$ $1 xx 10^21$ molecules are removed from $280 mg$ $280" mg"$ of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer: $8.4 \times 10^{- 3} moles$ $8.4 xx 10^- 3" moles"$ ]

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Answer 1 · 2018-06-13T17:16:00

Approximately $8.4$ $8.4$ millimoles

Explanation:

Total moles of Carbon monoxide ( $CO$ $"CO"$ ) is

$"n"_"Total" = (280 × 10^-3\ "g")/"28 g/mol" = 10^-2\ "mol"$

Moles of $"CO"$ removed is

$"n"_"removed" = 1 × 10^21 cancel"molecules" × "1 mol"/(6.022 × 10^23 cancel"molecules")$

$color(white)("n"_"removed") = 1.66 × 10^-3\ "mol"$

Moles of $"CO"$ left is

$"n"_"left" = "n"_"Total" - "n"_"removed"$

$color(white)("n"_"left") = 10^-2\ "mol" - (1.66 × 10^-3\ "mol")$

$color(white)("n"_"left") = (10 - 1.66) × 10^-3\ "mol"$

$color(white)("n"_"left") ≈ 8.4 × 10^-3\ "mol"$

Answer 2 · 2018-06-13T17:19:56

$8,34*10^-3$ moles

Explanation:

CO has a molecular mass of 28, so 280 mg are 10 millimoles.
$10^21$ molecules are $10^21/N_A =1.66*10^-3$ moles,
Subtracting $1.66*10^-3$ moles from $10*10^-3$ moles gives $8.34*10^-3$ moles

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$1 \times 10^{21}$ $1 xx 10^21$ molecules are removed from $280 mg$ $280" mg"$ of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer: $8.4 \times 10^{- 3} moles$ $8.4 xx 10^- 3" moles"$ ]

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1 xx 10^211×10211 xx 10^21 molecules are removed from 280" mg"280 mg280" mg" of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer:8.4 xx 10^- 3" moles"8.4×10−3 moles8.4 xx 10^- 3" moles"]

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Explanation:

Explanation:

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$1 \times 10^{21}$ $1 xx 10^21$ molecules are removed from $280 mg$ $280" mg"$ of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer: $8.4 \times 10^{- 3} moles$ $8.4 xx 10^- 3" moles"$ ]