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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?

Respuesta :

Neetoo

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

Cricetus

The density of the exhibit water will be "1.023 g/mL".

Density of water:

According to the question,

Water's volume, V = 5.00 mL

Empty beaker's weight = 18.926 g

Beaker's weight with water = 24.041 g

Firstly, the mass of water:

= Beaker's weight with water - Beaker's mass

= [tex]24.041-18.926[/tex]

= [tex]5.115[/tex]

hence,

The density be:

→ [tex]d = \frac{m}{v}[/tex]

By substituting the values,

     [tex]= \frac{5.115}{5.00}[/tex]

     [tex]= 1.023[/tex] g/mL

Thus the answer above is correct.

Find out more information about volumetric pipette here:

https://brainly.com/question/16973449

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