multistep conversions hw\nuse placement zeroes when needed (example: 0.045) and do not use commas (example…

multistep conversions hw\nuse placement zeroes when needed (example: 0.045) and do not use commas (example: 31500)\nfind the mass of $2.45 \\times 10^{24}$ molecules of $cbr_4$.\n$m = \\square$ g

multistep conversions hw\nuse placement zeroes when needed (example: 0.045) and do not use commas (example: 31500)\nfind the mass of $2.45 \\times 10^{24}$ molecules of $cbr_4$.\n$m = \\square$ g

Answer

Explanation:

Step1: Calculate moles from molecules

To find moles, use Avogadro's number ($N_A = 6.022\times 10^{23}$ molecules/mol). The formula is $n=\frac{N}{N_A}$, where $N = 2.45\times 10^{24}$ molecules.
$n=\frac{2.45\times 10^{24}}{6.022\times 10^{23}}\approx 4.07$ mol.

Step2: Find molar mass of $CBr_4$

Molar mass of C: $12.01$ g/mol, Br: $79.90$ g/mol.
Molar mass of $CBr_4 = 12.01 + 4\times79.90 = 12.01 + 319.6 = 331.61$ g/mol.

Step3: Calculate mass from moles

Use $m = n\times M$, where $n = 4.07$ mol, $M = 331.61$ g/mol.
$m = 4.07\times 331.61\approx 1349.65$ g.

Answer:

1349.65