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How many atoms are present in a 112 g sample of #CaBr_2#?

Answer 1

Noah Aldrich

#~~0.56# moles or #~~3.37*10^23# atoms

Molar mass of #CaBr_2# is #199.886# amu

Moles = #"Mass"/"Molar Mass"#

Here, the molar mass is 199.886 and the mass is 112g.

#:.# Moles of #CaBr_2# = #112/199.886~~0.56#

#1# mole = #6.02*10^23# atoms (Avogadro's number)

#:. 0.56# moles = #3.37*10^23# atoms

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Answer 2

Axel Alvarado

The first step in determining the number of atoms in an 112 g sample of CaBr2 is to calculate the molar mass of the mixture, which is 40.078 g/mol for calcium and 79.904 g/mol for bromine. This gives a total molar mass of 40.078 + (2 * 79.904) = 199.886 g/mol for CaBr2. Next, divide the given mass by the molar mass, which equals 112 g / 199.886 g/mol ≈ 0.560 moles. Since there are two moles of bromine atoms in a mole of CaBr2, multiply the number of moles by Avogadro's number, which is 6.022 x 10^23 atoms/mol, and then multiply by two to account for the two bromine atoms per molecule: 0.560 moles * 6.022 x 10^23 atoms/mol * 2 ≈ 6.77 x 10^23 atoms.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.