Ask question

Ask question

All categories

2 years ago

5

What is the difference between pure and applied chemistry?

1 answer:

bekas [8.4K]2 years ago

3 0

Answer:

Explanation:

The major difference between pure and applied chemistry is the purpose and intent of the study.

Pure chemistry deals with the study of matter, matter transformations, and interactions between the different materials of the world, for only the sake of gaining empirical knowledge about the various substances that exist in the world. It does not really seek to apply this knowledge to do anything industrial.

Applied chemistry is the study of chemistry with the aim of utilizing this knowledge to solve the various problems that man faces. This approach of study is not for knowledge sake alone, rather it is for industrial application

You might be interested in

This chart shows descriptions of the atoms for four different substances.

Whitepunk [10]

I think it’s Object 3
Hopefully this helps

8 0

2 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

2 years ago

A sample of magnesium oxide has a mass of 94.4 g. How many molecules

ella [17]

Answer: 1.414x10^24 molecules in 94.4g MgO

Explanation: molar mass MgO 40.204

molecules in 40.204 g MgO = avogadro number

molecules in 94.4 g MgO = (94.4/40.204)*avogadro number

(94.4/40.204)*6.02214076*10^23 = 14.14x10^23

7 0

2 years ago

प्रश्न 5. 'पुष्प की अभिलाषा' कविता में पुष्प के दवारा क्या अभिलाषा व्यक्त की गई

Kay [80]

Please ask in English

4 0

2 years ago

More pointss hahahahahahahaha

Oksana_A [137]

Answer:

thanks

Explanation:

fjnrjbn

7 0

2 years ago

Read 2 more answers