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sergey [27]
3 years ago
15

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall r

edox equation: 5 Ag (aq) +Mn2 (aq) +4H20()5Ag(s)+ MnO4 (aq) +8H' (aq)
a. 1
b. 2
c. 3
d. 5 (
e.?
Chemistry
1 answer:
Sergeu [11.5K]3 years ago
4 0
Answer:
             5 moles of electrons

Explanation:

The balance equation is as follow,

<span>                   5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O    →<span>    5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺

Reduction of Ag:

                         Ag⁺ + 1 e⁻    →    Ag
Or,
                         5 Ag⁺  +  5 e⁻    →    5 Ag

Oxidation of Mn:

                          Mn⁺²   →    MnO₄⁻  +  5 e⁻

Result:
          Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
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Ensure you write the states of the reactants and products.
slamgirl [31]
<h3>Answer:</h3>

Phosphoric acid reacts with magnesium hydroxide to produce magnesium phosphate and water via the following reaction:

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

(solid) (solid) (solid) (liquid)

<h3>Explaination:</h3>

This is a typical neutralization reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat,

ΔH < 0 , and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.

Doing so we obtain :

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

(solid) (solid) (solid) (liquid)

<h3>hope it helps :)</h3>
6 0
3 years ago
Read 2 more answers
You have a balloon containing 1 L of air at STP in a vacuum chamber. What will the volume of the balloon be when you reduce the
Ipatiy [6.2K]
P1V1/T1=P2V2/T2   P2=1/2 P1 V1=1 T1=298K

1 P1/298= (1/2) P1V2/373  cross P1

1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L


5 0
3 years ago
The question is below
rodikova [14]
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
2Fe(3+) + 2e-  ----> 2 Fe(2+)                2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0
3 years ago
The type of radiation which is identical to a high energy electron is known as a(n). beta, alpha, or gamma particles
Schach [20]
The type of radiation which is identical to a high energy electron is known as a(n) beta.

Answer: A) or the first option.
5 0
3 years ago
Read 2 more answers
The following compound has been found effective in treating pain and inflammation (J. Med Chem. 2007, 4222). Which sequence corr
love history [14]
<h3><u>Full Question:</u></h3>

The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?

A) 1 < 2 < 3

B) 2 < 3 < 1

C) 3 < 1 < 2

D) 1 < 3 < 2

<h3><u>Answer: </u></h3>

The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.

Option B

<h3><u>Explanation. </u></h3>

Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.

A carbonyl compound is a sp^2 hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.

In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.

7 0
3 years ago
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