How would you the mass in grams of $4.52 \times 10^{- 3}$ $4.52 xx10^-3$ $C_{20} H_{42}$ $"C"_20"H"_42"$ ? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.

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How would you the mass in grams of $4.52 \times 10^{- 3}$ $4.52 xx10^-3$ $C_{20} H_{42}$ $"C"_20"H"_42"$ ? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.

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Answer 1 · 2017-12-23T03:35:36

Take the product... $molar quantity \times molar mass$ $"molar quantity"xx"molar mass"$ ; I gets a tad over $1 \cdot g$ $1*g$ ...

Explanation:

You got the molar quantity... $4.52 \times 10^{- 3} \cdot m o l$ $4.52xx10^-3*mol$ ...(you have not specified the units in your question, but I assume that these units pertain); and multiply these units by the molar mass...

$4.52 \times 10^{- 3} \cdot m o l \times (20 \times 12_{atomic mass of carbon} + 42 \times 1_{atomic mass of hydrogen}) \cdot g \cdot m o l^{- 1} = ? ? \cdot g$ $4.52xx10^-3*molxx(20xx12_("atomic mass of carbon")+42xx1_("atomic mass of hydrogen"))*g*mol^-1=??*g$

Answer 2 · 2017-12-23T06:50:07

The mass of $4.52 \times 10^{- 3} {mol C}_{20} H_{42}$ $4.52xx"10"^(-3) "mol C"_20"H"_42"$ is $1.27 g$ $"1.27 g"$ .

Explanation:

If the mass of $4.52 \times 10^{- 3} {mol C}_{20} H_{42}$ $4.52xx"10"^(-3) "mol C"_20"H"_42"$ is what you are looking for:

Determine the molar mass of $C_{20} H_{42}$ $"C"_20"H"_42"$ .

$Molar mass$ $"Molar mass"$ $.$ $color(white)(.)$ $C_{20} H_{42} :$ $"C"_20"H"_42:$ $(20 \times 12.0 g/mol C) + (42 \times 1.0 g/mol H) = {282 g/mol C}_{20} H_{42}$ $(20xx"12.0 g/mol C")+(42xx"1.0 g/mol H")="282 g/mol C"_20"H"_42"$

Mass of $4.52 \times 10^{- 3} {mol C}_{20} H_{42}$ $4.52xx"10"^(-3) "mol C"_20"H"_42"$

Multiply mol $C_{20} H_{42}$ $"C"_20"H"_42"$ by its molar mass.

$4.52xx"10"^(-3) color(red)cancel(color(black)("mol C"_20"H"_42))xx(282"g C"_20"H"_42)/(1color(red)cancel(color(black)("mol C"_20"H"_42)))="1.27 g C"_20"H"_42"$ rounded to three significant figures

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How would you the mass in grams of $4.52 \times 10^{- 3}$ $4.52 xx10^-3$ $C_{20} H_{42}$ $"C"_20"H"_42"$ ? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.

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Explanation:

Explanation:

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How would you the mass in grams of 4.52 xx10^-34.52×10−34.52 xx10^-3 "C"_20"H"_42"C20H42"C"_20"H"_42"? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.

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Explanation:

Explanation:

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How would you the mass in grams of $4.52 \times 10^{- 3}$ $4.52 xx10^-3$ $C_{20} H_{42}$ $"C"_20"H"_42"$ ? Use the molar masses: C = 12.0 g/mole; H = 1.0 g/mol.