Explain how to calculate the density of a regular shaped object such as a rectangular box.

Chemistry

1 answer:

belka [17]1 year ago

4 0

The density of a regular shaped object such as a rectangular box is calculated as thus:

The amount of matter is its mass, and the space it takes up is its volume.

So therefore, to calculate the density of an object such as the a regular shaped one like a rectangular box;

Its density = mass / volume.

<h3>Derived quantities</h3>

These are physical quantities which are obtained from fundamental quantities. I'm order words, derived quantities are quantities which are obtained from the combination of of fundamental quantities either by addition, division or multiplication

Some few examples of derived quantities are as follows:

Density

Velocity
Upthrust
Volume
Acceleration

So therefore, the density of a regular shaped object such as a rectangular box is calculated as thus:

The amount of matter is its mass, and the space it takes up is its volume.

So therefore, to calculate the density of an object such as the a regular shaped one like a rectangular box;

Its density = mass / volume.

Learn more about derived quantities:

brainly.com/question/28180944

#SPJ1

You might be interested in

Liquid helium boils at –268.93 °C. What is the boiling point of helium on the Kelvin temperature scale?

BARSIC [14]

In order to change celcius to kelvin always add 73 to it leaving you with -195.93

4 0

2 years ago

Read 2 more answers

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$