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Chapter 3: Problem 140

In the spring of \(1984,\) concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecular formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) and until 1984 was commonly used as a plant fumigant. The federal limit for \(\mathrm{EDB}\) in finished cereal products is 30.0 parts per billion (ppb), where \(1.0 \mathrm{ppb}=\) \(1.0 \times 10^{-9} \mathrm{g}\) of \(\mathrm{EDB}\) for every 1.0 \(\mathrm{g}\) of sample. How many molecules of EDB are in 1.0 lb of flour if 30.0 ppb of EDB is present?

Short Answer

Expert verified
There are approximately \(6.4 \times 10^{15}\) EDB molecules in 1.0 lb of flour when 30.0 ppb of EDB is present.

Step by step solution

01

Convert the weight of flour from lb to grams

1 lb is equal to 454 grams. So, 1.0 lb of flour is equivalent to 454.0 grams.
02

Calculate the weight of EDB in 1.0 g of flour

30.0 ppb means there are \(1.0 \times 10^{-9}\) g of EDB per 1.0 g of sample. Therefore, for 1.0 g of flour, there are \(30.0 \times (1.0 \times 10^{-9})\) g of EDB.
03

Determine the total weight of EDB in the entire sample of flour

Now we need to find out the total weight of EDB in the given flour sample. Since we have 454.0 grams of flour, the total weight of EDB will be: \(EDB_{total} = 454.0 \times 30.0 \times (1.0 \times 10^{-9})\) g
04

Find the number of EDB moles using its molecular weight

The molecular formula of EDB is \(C_2H_4Br_2\). Using the molecular weights of carbon (C), hydrogen (H), and bromine (Br), we can calculate the molecular weight of EDB: Molecular weight of EDB = \((2 \times 12.01) + (4 \times 1.007) + (2 \times 79.90)\) g/mol Now, we can calculate the number of moles of EDB present in the flour sample using the formula: Moles of EDB = \(\frac{EDB_{total}}{Molecular\:weight\:of\:EDB}\)
05

Calculate the number of EDB molecules using Avogadro's number

Avogadro's number (N) is approximately \(6.022 \times 10^{23}\). We can calculate the number of EDB molecules using the formula: Number of EDB molecules = Moles of EDB × Avogadro's number After performing all the calculations, you will get the number of EDB molecules in the given sample of flour.

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Most popular questions from this chapter

Hexamethylenediamine $\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)$ is one of the starting materials for the production of nylon. It can be prepared from adipic acid $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)$ by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide $\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)$ could be produced? $\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad$ (unbalanced)

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$ 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow $$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains 91.27\(\% \mathrm{E}\) and 8.73\(\% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8},\) calculate the atomic mass of \(\mathrm{E}\)

Balance the following equations: a. $\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{S}_{3}(s)$ b. \(\operatorname{NaHCO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) $\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ c. \(\quad \mathrm{KClO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) \(\mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. $\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)$
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