In the following redox reaction, the reducing agent is MnO2 (option D). Details about reducing agent can be found below.
<h3>What is a reducing agent?</h3>
A reducing agent in a redox reaction is any substance that reduces, or donates electrons to another, hence, it becomes oxidized.
According to this question, a redox reaction is given as follows: MnO2(s) + 4H+(aq) + 2Cl–(aq) = Mn2+(aq) + 2H2O(l) + Cl2(g)
As shown in the equation, MnO2 is oxidized into Mn2+, therefore, it is the reducing agent.
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Explanation:
A stable isotope is one that does not emit radiation, or, if it does its half-life is too long to have been measured. It is believed that the stability of the nucleus of an isotope is determined by the ratio of neutrons to protons.
Hope this helps you out : D
The region where an electron is most likely to be located is known as the electron cloud.
Protons and neutrons are found inside the nucleus
The fomula is NH4 (1+)
There are only two elements N and H.
As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.
N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.
You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.
In conclusion the oxidation state of H in NH4 (1+) is 1+.
Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.
That implies that 4* (1+) + x = 1+
=> x = (1+) - 4(+) = 3-
Answer: the oxidation state of N is 3-, that is the option b.
