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n200080 [17]
2 years ago
6

Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta

ining species in the product is HCO3–(aq). Add H2O and H+ to balance the H and O atoms in the equation. Do not add electrons; you may leave the half-reaction unbalanced with respect to charge.
Chemistry
1 answer:
Leokris [45]2 years ago
3 0

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

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Which of the following shows a Bronsted-Lowry acid reacting?
7nadin3 [17]
<h3>Answer:</h3>

         Option-C:  HCl + H₂O  →   H₃O⁺ + Cl⁻

Explanation:

       Bronsted-Lowery concept of Acid and Base defines Acid as that specie which tends to donate H⁺ (Hydrogen Ion) and bases are those species which accepts H⁺ from Acids.

In selected option, HCl is reacting as Acid as it donates H⁺ to water (lowery bronsted base).

Also, the correspong acid is converted into conjugate base (i.e. Cl⁻) and base is converted into conjugate acid (i.e. H₃O⁺)

8 0
2 years ago
Read 2 more answers
For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

3 0
2 years ago
A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container. The carbon dioxide took 1.25 times as lon
Aleks04 [339]

Answer:

CO

Explanation:

From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.

t1/t2=√m1/m2

Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g

1.25/1=√44/m2

Therefore m2=28g CO

7 0
2 years ago
C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti
Andreas93 [3]

Answer :

Ethanol (C_2H_5OH) act as reducing agent.

The smallest possible integer coefficient of MnO_4^- in the combined balanced equation is, 4

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

C_2H_5OH(aq)+MnO_4^-(aq)\rightarrow CH_3COOH(aq)+Mn^{2+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : C_2H_6O\rightarrow C_2H_4O_2

Reduction : MnO_4^-\rightarrow Mn^{2+}

  • Now balance oxygen atom on both side.

Oxidation : C_2H_6O+H_2O\rightarrow C_2H_4O_2

Reduction : MnO_4^-\rightarrow Mn^{2+}+4H_2O

  • Now balance hydrogen atom on both side.

Oxidation : C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^+

Reduction : MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O

  • Now balance the charge.

Oxidation : C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^++4e^-

Reduction : MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : 5C_2H_6O+5H_2O\rightarrow 5C_2H_4O_2+20H^++20e^-

Reduction : 4MnO_4^-+32H^++20e^-\rightarrow 4Mn^{2+}+16H_2O

The balanced chemical equation in acidic medium will be,

5C_2H_6O+4MnO_4^-+12H^+\rightarrow 5C_2H_4O_2+4Mn^{2+}+11H_2O

In the redox reaction ethanol act as reducing agent and permanganate ion act as an oxidizing agent.

4 0
2 years ago
Methane (CH4) reacts with excess oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). What is the percent yield of c
musickatia [10]
(29.8 g) / [0.184 mol (44.00964 g CO2/mol)] =0.832= 83.2% yield CO2

(hope this helps)
4 0
2 years ago
Read 2 more answers
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