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3 years ago

What is the difference between applied and pure chemistry?

Chemistry

1 answer:

My name is Ann [436]3 years ago

5 0

I hope this helps.
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Help fast. . Zz. Zz z z z

Natasha2012 [34]

Answer:

step c is filtration

step b is decantation

6 0

3 years ago

Which of the following represents the correct ranking in terms of increasing boiling point? A. n-butane < 1-butanol < diet

bonufazy [111]

Answer:

The answer to your question is: letter E

Explanation:

Normally, the correct order of boiling points is:

Alcohols > Ketones > Ether > Alkane

Then

A. n-butane < 1-butanol < diethyl ether < 2-butanone

B. n-butane < 2-butanone < diethyl ether < 1-butanol

C. 2-butanone < n-butane < diethyl ether < 1-butanol

D. n-butane < diethyl ether < 1-butanol < 2-butanone

E. n-butane < diethyl ether < 2-butanone < 1-butanol

(- 1°C) < 34.6°C < 79.64°C < 117.7°C

6 0

3 years ago

Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at

Svetllana [295]

Answer:

The rate at which $P_4$ is being produced is 0.0228 M/s.

The rate at which $PH_3$ is being consumed is 0.0912 M/s.

Explanation:

$4PH_3\rightarrow P_4(g)+6H_2(g)$

Rate of the reaction : R

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which hydrogen is being formed = $\frac{d[H_2]}{dt}=0.137 M/s$

$R=\frac{1}{6}\frac{d[H_2]}{dt}$

$R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s$

The rate at which $P_4$ is being produced:

$R=\frac{1}{1}\frac{d[P_4]}{dt}$

$0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which $PH_3$ is being consumed :

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}$

$0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}$

$\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s$

6 0

3 years ago

Convert 2.50 x 10^5 seconds into years

ra1l [238]

0.00792219116 years

...........................

8 0

3 years ago

A sprinter set a high school record in track and field, running 200.0 m in 21.7 s . What is the average speed of the sprinter in

Yuliya22 [10]

Given data:

Distance travelled (d)= 200.0 m

Time taken (t) = 21.7 s

To determine:

The average speed in km/hr

Calculation:

Convert distance from m to km

1000 m = 1 km

200.0 m = 1 km * 200.0 m/1000 m = 0.2 km

Convert time from sec to hours

3600 sec = 1 hour

21.7 sec = 1 hour * 21.7 sec/3600 sec = 6.027 *10^-3 hrs

Speed is defined as the amount of distance travelled per unit time

Speed = distance/time = 0.2 km/6.027*10^-3 hr

= 33.18 km/hr (or 33.2 km/hr)

7 0

3 years ago