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3 years ago

While rowing in a race, John pulls the handle of an oar 0.80 m on each

Chemistry

1 answer:

Korvikt [17]3 years ago

8 0

Answer:

Stroke D (Output Distance) = 0.533 meter (Approx)

Explanation:

Given:

Stroke D (Input Distance) = 0.80 meter

Mechanical advantage (MA) = 1.5

Find:

Stroke D (Output Distance) = ?

Computation of Stroke D (Output Distance) :

⇒ Stroke D (Output Distance) = Stroke D (Input Distance) / Mechanical advantage (MA)

⇒Stroke D (Output Distance) = 0.80 meter / 1.5

⇒ Stroke D (Output Distance) = 0.533 meter (Approx)

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

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Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

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By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

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Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

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