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pav-90 [236]
3 years ago
5

What is the difference between pure and applied chemistry?

Chemistry
1 answer:
bekas [8.4K]3 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

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A soccer ball is moving to the right. how can you increase the velocity of the ball?
Fantom [35]

Answer: The air will move more quickly around one side, making less pressure on that side of the ball

Explanation:

3 0
2 years ago
Select the reagents you would use to synthesize the compounds below from benzene. Use the minimum number of steps. No more than
jekas [21]

Answer:

Explanation:

Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.

8 0
3 years ago
What were the limitations of Newlands Law of Octaves.?<br><br> No spam ❌❌<br> or else (-_-メ)​
liubo4ka [24]

Answer:

The major limitations of Newlands' law of octaves were : (i) It was applicable to only lighter elements having atomic masses upto 40 u, i.e., upto calcium. After calcium, the first and the eighth element did not have similar properties

5 0
3 years ago
Read 2 more answers
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine so
sergeinik [125]

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

<em>Moles C2H5NH2:</em>

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

<em>Moles HNO3:</em>

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

6 0
3 years ago
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