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vekshin1
2 years ago
13

Which method is adopted in the seperation of lead chloride from water​

Chemistry
1 answer:
Sergio039 [100]2 years ago
3 0

Explanation:

by dissolving the mixture of lead sulphate and lead chloride in water we can separate the two. after dissolving the mixture on water lead sulphate can be obtained as the solid that 's left behind lead chloride can be recovered by evaporating.

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Where did they start water conservation in tamil nadu . short answer
OleMash [197]

Explanation:

The 2019 water crisis in Chennai has made us realize the importance of saving water more than ever. Water, as we all know, is a finite resource without which our planet would be a barren wasteland. Today with our increasing population it would be logical to say that our water consumption has also increased. And with increasing demand and lesser supply, water scarcity arises. Our ancestors who had foreseen the potential dangers of water scarcity had developed methods to conserve water that was suited for the varied terrain of the Tamil-speaking kingdoms.

Traditional Rainwater Conservation methods of Tamil Nadu

Eri

There are no perennial rivers in Tamil Nadu except the Thamirabharani River which flows through Thirunelveli district. And so, several hundred years ago a simple system was devised to utilize the rainwater to the fullest. An Eri or tank system is one of the oldest forms of water conservation systems in India. Many Eris are still in use in Tamil Nadu and play an active role in irrigation. They act as water reservoirs and flood control systems. They prevent soil erosion, recharge groundwater, and prevent wastage of runoff water during heavy rainfall.

Kudimaramathu

Kudimaramathu is one of the old traditional practice of stakeholders participating in the maintenance and management of irrigation systems. During earlier days, citizens of a village used to actively participate in maintaining the water bodies of their village by deepening and widening the lakes and ponds and restoring the water bodies back to their original form. The silt, rich in nutrients, collected in the process would be used by the farmers themselves in their field. A sense of collective ownership ensured the continued survival of the water bodies.

6 0
2 years ago
How is the melting point of a substance recorded
GrogVix [38]

1. slowly heat stubstance.

2. once the substance is at the most liqued state take the temp. that's the melting point of that subtance.

hope that helps, any other questions feel free to DM me dont wate your points. :)

6 0
2 years ago
What are the answers please help
LenaWriter [7]
Thank you for the free 15 points .
4 0
2 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135

8 0
3 years ago
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