The oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.
<h3>What is an oxidizing and reducing agent?</h3>
An oxidizing agent is any substance that oxidizes, or receives electrons from another substance and as a result, becoming reduced.
On the other hand, a reducing agent is any substance that reduces or donates electrons to another and as a result becomes oxidized.
According to this reaction; H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)
- H2S accepts electrons from Cl2 and becomes reduced to S
- Cl2 donates electrons to H2S and becomes oxidized to HCl
Therefore, the oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.
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Answer:
2H2(g]+O2(g]→2H2O(l]]. Notice that the reaction requires 2 moles of hydrogen gas and 1
Explanation:
The molar mass, in grams, of a mole for an element is equal to the atomic weight if the element.
Alka-seltzer in an antacid that contains a mixture of sodium bicarbonate and citric acid. When the tablet is dissolved in water, the reactants which are in solid form in tablet become aqueous and react with each other.
During this reaction, Carbon Dioxide gas is evolved which causes the reaction mixture to fizz. The equation is given below.
Rate of the above reaction is affected by the Temperature.
As the temperature increases , the rate of the reaction increases. This happens because at higher temperature, the collisions between reacting species are more which result in formation of product in less time. This increases the rate of reaction.
We have been given equal volumes of water for each beaker. But the temperature of beaker c is 80°C which is the highest temperature. That means the reaction in beaker c is fastest.
Whereas beaker a is at lowest temperature (30°C) , therefore the reaction in beaker a would be slowest .
Therefore the answer that correctly orders the reaction rates from fastest to slowest reaction is beaker c > beaker b > beaker a
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!
