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ivolga24 [154]
2 years ago
14

What is the mass (in grams) of 9.93 Ă— 1024 molecules of methanol (CH3OH)?

Chemistry
1 answer:
WINSTONCH [101]2 years ago
7 0

Answer is: mass of methanol is 528.32 grams.

N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.

n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).

n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.

n(CH₃OH) = 16.49 mol; amount of substance.

m(CH₃OH) = n(CH₃OH) · M(CH₃OH).

m(CH₃OH) = 16.49 mol · 32.04 g/mol.

m(CH₃OH) = 528.32 g.

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azamat

Answer:

(a) m_{gold}=7.322g

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Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

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\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

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