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Marizza181 [45]

3 years ago

6

A graduated cylinder was filled with 25.0 mL of water. A solid object was immersed in the cylinder, raising the water level to 3

2.9 mL. The volume of the object was calculated as 7.9 mL. The solid has a mass of 69.2 grams. What is the density of the object?

1 answer:

yawa3891 [41]3 years ago

7 0

Answer:

Density of the object is 8759.494 grams/L

Explanation:

As we know density of an object is mass of the object divided by its volume

Given

Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.

Hence, volume of object is equal to

$32.9 -25 = 7.9$ mL

Mass of the object is $69.2$ grams

Density

$= \frac{69.2}{7.9 *10^{-3}}\\= 8759.494$ grams/L

Density of the object is 8759.494 grams/L

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How do you calculate mass using density and volume?

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Example:

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3 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

3 years ago