All categories

4 years ago

What's the differentiating characteristic between an element and a compound

Physics

2 answers:

IRISSAK [1]4 years ago

8 0

Answer:

an element is a pure substance whereas a compound is a mixture of two or more elements.

umka21 [38]4 years ago

6 0

The basic difference between a compound and an element is its constituents, in which the element is a pure individual chemical substance which has similar atom and compound is a mixture of two or more elements bounded chemically.

Explanation:

Elements:

Elements are pure homogenous substance.
Elements have similar atoms grouped together.
Element is being represented with the chemical scientific symbol.

Compounds:

Compound are heterogeneous mixture of elements.
Compound has different atoms which is combined in fixed ratio.
Compound is given by chemical formula.

You might be interested in

what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s

kotegsom [21]

Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.

7 0

3 years ago

What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An

GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}$

$E=\frac{2m_{0}c^}2}{\sqrt{3}}$

$E=\frac{2\times0.001\times9\times10^{16}}{1.732}$

$E=1.04\times10^{14}J$

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, $E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV$

4 0

3 years ago

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago

How many different values of l are possible for an electron with principal quantum number n=5?

lions [1.4K]

Answer:

different value to l if n=5 are 0,1,2,3,4

7 0

3 years ago

Which of the following is a terrestrial habitat? A) Pond B) Garden C) Lake D) River

kenny6666 [7]

Answer:

garden

Explanation: All the other habitats are aquatic

8 0

3 years ago