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Musya8 [376]
1 year ago
5

According to the beer-lambert law, what happens to the absorbance value of a light-absorbing chemical when its concentration inc

reases?.
Chemistry
1 answer:
Nana76 [90]1 year ago
5 0

According to the beer-lambert law when its concentration increases  so does the absorbance increase.

beer-lambert law states that the amount of energy absorbed or transferred by a solution is proportional to the molar absorptivity of the solution and the concentration of the solute.

This means that concentrated solutions absorb more light than dilute solutions.

so the According to the beer-lambert law concentration increases, so does the absorbance. Therefore, absorbance is directly proportional to concentration.

learn more about beer-lambert law at brainly.com/question/8831959

#SPJ4

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What is the name of(NH4)3PO4
Tju [1.3M]

Answer:

What is the name of the compound (NH4) 3PO4? (NH)4+ is an ammonium radical and (PO4) 3- is a phospate radical. When we start writing the compound,the valency of phosphate goes to ammonium as subscript,and that of ammonium goes to phosphate. Hence the formula (NH4)3 PO4 and it's name is ammonium phospate.

Explanation:

5 0
3 years ago
Which of the following representations shows the correct placement of trophic levels?
OleMash [197]
You have to start listing from the bottom :

3. Secondary Consumers

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1. Producer

4 0
3 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
based on the techniqes yo have learned in the organic chemistry lab how would you seperate any unreated alcohol from the ester
mihalych1998 [28]

Answer:

Fractional distillation and HP-LC

Explanation:

This is a technique useful for analytes with close boiling points. Any alcohol-ester azotopes can be further refined using high-performance liquid chromatography (HP-LC) column.

3 0
3 years ago
If you are given 12M HCL how would you make a 6M of 100mL? SHow the math and explain the process
AlladinOne [14]

dilution

V₁M₁=V₂M₂

V₁.12 = 100.6

V₁=50 ml

Add water 50 ml to 50 ml 12 M

7 0
2 years ago
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