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The Atomic Mass of boron is 10.811 while that of bromine is 79.904. Thus the ratio of the mass of bromine to the total mass of boron tribromide must be:

3(79.904)/[3(79.904) + 10.811] = 0.9568. The required mass of BBr3 thus is 11.8/0.9568 = 12.3 g, to the justified number of significant digits.

Q: How many grams of boron tribromide must be weighed out to get a sample of boron tribromide that contains 11.8 g of bromine?

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Bromine is an element, and no other kinds of atoms or elements are present with it, in a pure sample.

Which sample contains the greatest number of atoms. A sample of Mn that contains 3.29E+24 atoms or a 5.18 mole sample of I?The sample of _____ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of I to atoms of I.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.18 moles of I to atoms of I:atoms I= 5.18 mol I6.02 x 1023 atoms I = 3.12E+24 atoms I1 mol IMultiply by atoms per mole. Moles cancel out.The sample of Mn contains 3.29E+24 atoms.Since 3.12E+24 is smaller than 3.29E+24, the sample of Mn contains the greatest number of atoms.

150 (50 x 3)

This depends on the mass of the gold sample.

Either that or wear gloves. Your trying to avoid direct contact.

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There isn't a direct correlation. Bromine Number is a measure of the amount of olefins in a sample. Diene Valve is a measure of the conjugated diolefins in a sample. In hydrocarbons, samples with high Diene Value also tend to have high Brominie numbers as a result of processing in a hydorgen deficient atmosphere.

Amount of Br2 = mass of sample / molar mass = 160 / 2(79.9) = 1.00mol

Contains a sample of "Cavern" by Liquid Liquid

For example to contain a sample to be weighed; or to make a chemical reaction at small scale; or to cover a beaker.

the leg

Which sample contains the greatest number of atoms. A sample of Mn that contains 3.29E+24 atoms or a 5.18 mole sample of I?The sample of _____ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of I to atoms of I.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.18 moles of I to atoms of I:atoms I= 5.18 mol I6.02 x 1023 atoms I = 3.12E+24 atoms I1 mol IMultiply by atoms per mole. Moles cancel out.The sample of Mn contains 3.29E+24 atoms.Since 3.12E+24 is smaller than 3.29E+24, the sample of Mn contains the greatest number of atoms.

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