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krek1111 [17]
2 years ago
6

Describe a method of Municipal treatment of water with neat diagram?​

Chemistry
1 answer:
stich3 [128]2 years ago
4 0
Water treatment process
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What happen to the energy that is lost when water change to gas
Y_Kistochka [10]

Answer:

This process, which is the opposite of vaporization, is called condensation. As a gas condenses to a liquid, it releases the thermal energy it absorbed to become a gas. During this process, the temperature of the substance does not change. The decrease in energy changes the arrangement of particles.

6 0
1 year ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
Can someone help me balance this equation please?<br> Br2 + S2O32– + H2O → Br1– + SO42– + H+
tamaranim1 [39]

Explanation:

Br2 + S2O32- + 5H2O –> 2Br- + 2SO4 + 10H+ + 6e

7 0
2 years ago
A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution to produce a precipi
OLEGan [10]

Answer:

First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of HNO_{3}.

Explanation:

In presence of NaCl, AgNO_{3} forms an insoluble precipitate of AgCl.

Reaction: Ag^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)

In presence of NH_{3}, AgCl gets dissolved into solution due to formation of soluble [Ag(NH_{3})_{2}]^{+} complex.

Reaction: AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)

In presence of HNO_{3}, [Ag(NH_{3})_{2}]^{+} complex gets destroyed and free Cl^{-} again reacts with free Ag^{+} to produce insoluble AgCl

Reaction: [Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)

4 0
3 years ago
A gas has a temperature of 273.15 k and a pressure of 101.325 kpa. what can be concluded about the gas?
FromTheMoon [43]
We are given with a <span>temperature of 273.15 Kelvin and a pressure of 101.325 kPa. Using PV=nRT rule of ideal gas, where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature, we can conclude that the conditions are under STP(standard temp and pressure) and one mole of this gas occupies 22.4 L of volume.</span>
5 0
2 years ago
Read 2 more answers
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