Which element is reduced in the reaction below? A. Na B. CI C. Ba D. None
2 answers:
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
For the given chemical reaction:
The above chemical reaction is considered as double displacement reaction because here exchange of ions takes place.
In these type of reactions, oxidation state of the atoms remains the same.
<u>On the reactant side:</u>
Oxidation state of barium = +2
Oxidation state of chlorine = -1
Oxidation state of sodium = +1
<u>On the product side:</u>
Oxidation state of barium = +2
Oxidation state of chlorine = -1
Oxidation state of sodium = +1
Thus, the correct answer is Option D.
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Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer: is a reactant; it is present before the reaction occurs.
Explanation:
In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.
Thus and HCl are reactants here whereas
,
and
are products.