This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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Answer:
a. +2
b. +3
c. -1
Explanation:
The typical oxidation states can be determined from the periodic table based on the number of valence electrons an atom has.
a. Calcium belongs to group 2A, meaning it has 2 valence electrons and, therefore, would have an oxidation state of +2 in compounds.
b. Aluminum is in group 3A, meaning it has 3 valence electrons and would have an oxidation state of +3 in compounds when the 3 electrons are lost.
c. Fluorine would become fluorine if it gained 1 additional electron to achieve an octet, so its oxidation state would be -1.
An organisam is part of your body plant life
Explanation:
Formula to calculate how many particles are left is as follows.
N = 
where, 
= number of initial particles
l = number of half lives
As it is given that number of initial particles is 
and number of half-lives is 3.
Hence, putting the given values into the above equation as follows.
N = 
= 
= 
or, 
Thus, we can conclude that particles of radioactive nuclei remain in the given sample.
In five hours we've gone through 5 half lives so the answer is:
particles
Answer:
The new concentration is 2.03M
Explanation:
Step 1: Data given
A 200 mL 3.55 M HBr is diluted with 150 mL
Step 2: The dilution
In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.
Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume
In this case, the volume of the stock solution is 200 mL
Adding 150 mL of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL
The dilution factor wll be 350/200 = 1.75
This makes the diluted concentration:
3.55/1.75 = 2.03M
The new concentration is 2.03M
