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mars1129 [50]
3 years ago
9

In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Chemistry
1 answer:
Oksanka [162]3 years ago
8 0

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

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Where did they start water conservation in tamil nadu . short answer
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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
3 years ago
The reaction of hydrogen gas with oxygen gas is what type of reaction precipitation single replacement
Oksi-84 [34.3K]
The reaction of hydrogen gas (H2) with oxygen gas (O2) is a COMBINATION or SYNTHESIS reaction, because multiple substances combine to form fewer substances. Here, the two gases form one substance, water (H2O):
2H2 + O2 -- > 2H2O
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3 years ago
Convert 3.9x10^5mg to dg
bonufazy [111]
Thank you for posting your math problem here. To convert 3.9x10^5mg to dg the answer is <span>3.9 x 10^3 dg. Below is the solution: 

Solution:

</span><span>1mg=0.01dg 
</span><span> dg= 3.9 X 10^5mg
</span>dg = <span>(3.9 X 10^5) x 0.01
dg = </span><span>3.9 x 10^3 </span>
7 0
3 years ago
Read 2 more answers
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

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