I believe it might be A, hope this helps!
Answer: 1.
: oxidation reduction
2.
: precipitation
3.
: Double displacement
Explanation:
Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Double displacement reaction is defined as the reaction where exchange of ions takes place. Double displacement reaction in which one of the product remain in solid form are represented by (s) after their chemical formulas. Such double displacement reaction are called as precipitation reaction.

Double displacement reaction is defined as the reaction where exchange of ions takes place.

Single displacement reaction is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.
Decomposition reaction is defined as the reaction where a single substance breaks down into two or more simpler substances.
Synthesis/Combination reaction is defined as the reaction where substances combine in their elemental state to form a single compound.
The true statement is that after reaching equilibrium, the rate of forming products and reactants is the same.
<h3>What is true about the given reaction?</h3>
The given reaction shows a reaction between A and B to form CD
The reaction is a reversible reaction.
A reversible reaction is a reaction which can proceed in either of two ways where the reactants can react to form the product and also the products an break down to form the reactants.
In the reaction given, as the concentration of A and b decreases, the concentration of CD increases and vice versa.
At equilibrium, the rate of formation of CD is equal to the the rate of decomposition of CD.
Therefore, the true statement is that after reaching equilibrium, the rate of forming products and reactants is the same.
In conclusion, a reaction at equilibrium has the forward and backward reactions occurring at the sane rate.
Learn more about equilibrium reaction at: brainly.com/question/18849238
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Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.