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enyata [817]
2 years ago
8

A team adds 5.00 mL of exhibit water to an empty beaker using a volumetric pipette. What is the density of the exhibit water if

the empty beaker weighed 18.926g and the beaker weighed 24.041g after adding the water sample?
Chemistry
1 answer:
Law Incorporation [45]2 years ago
7 0

Answer:

d = 1.023 g/mL

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

Volume of water = 5.00 mL

Weight of empty beaker = 18.926 g

Weight of beaker with water = 24.041 g

Density of water = ?

Solution:

First of all we will calculate the mass of water.

Mass of water = mass of beaker with water - mass of beaker

Mass of water = 24.041 g - 18.926 g

Mass of water = 5.115 g

Density:

d = m/v

d = 5.115 g/ 5.00 mL

d = 1.023 g/mL

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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
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Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

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According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

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Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

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{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

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