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aksik [14]
4 years ago
13

For the following reaction, identify the element that was oxidized, the element that was reduced, and the oxidizing agent. Give

an explanation for each answer.
Zn + H2SO4 ZnSO4 + H2
Chemistry
2 answers:
Thepotemich [5.8K]4 years ago
8 0

Answer: Element Zn is being oxidized, Element hydrogen is being reduced and it the oxidizing agent.

Explanation:

Oxidation reaction is defined as the reaction in which an element looses its electrons. The oxidation state is increased in these type of reaction.The element undergoing these reactions are said to be oxidized.

Reduction reaction is the reaction in which an element accepts electrons. The oxidation state decreases in these type of reactions. The element undergoing these reactions are said to be reduced.

Oxidizing agents are defined as the agents which helps in the oxidation of other element and itself gets reduced.

Reducing agents are defined as the agents which helps in the reduction of other element and itself gets oxidized.

For the given reaction:

Zn+H_2SO_4\rightarrow ZnSO_4+H_2

Zinc is getting oxidized to Zn^{2+}

Hydrogen is getting reduced to H_2

Oxidation reaction: Zn\rightarrow Zn^{2+}+2e^-

Reduction reaction: 2H^++2e^-\rightarrow H_2

Here, hydrogen is considered as the reducing agent.

Thepotemich [5.8K]4 years ago
3 0
Oxidation is the Loss of electron 
Reduction is the Addition of Electron
<span>Zn + H2SO4  ------>  ZnSO4 + H2

</span>Zn(+2)/Zn ----------------------)       Zn ----------->Zn(+2) + 2e-
H(+)/H2 ------------------------)        2H(+) + 2e -------------> H2
So ,
 
Zn is the reducing agent and H+ is the oxidation agent
Therefore,
Zn is oxidized and H+ is reduced
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A hydraulic system is designed to lift cars for inspection in a service station. The narrow end of the system has a surface area
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<u>Given:</u>

Surface area at the narrow end, A1 = 5.00 cm2

Force applied at the narrow end, F1 = 81.0 N

Surface area at the wide end, A2 = 725 cm2

<u>To determine:</u>

Force F2 applied at the wide end

<u>Explanation:</u>

Use the relation

F1/A1 = F2/A2

F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N

Ans: (b)

The force applied at the wide end = 11,745 N

5 0
3 years ago
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In a data set, the number that appears the most is called the ( ).
scZoUnD [109]

Answer:

Mode

Explanation:

The mode is the number that appears most frequently in a data set. A set of numbers may have one mode, more than one mode, or no mode at all. Other popular measures of central tendency include the mean, or the average (mean) of a set, and the median, the middle value in a set.

4 0
4 years ago
For a single component system, why do the allotropes stable at high temperatures have higher enthalpies than allotropes stable a
solniwko [45]

Answer:

The difference in the magnetic orientation influences the thermal stability of the allotropes of iron.

Explanation:

It is known that the allotropes of iron exist in three phases: α - phase, β- phase, and γ-phase. However, two prominent structures are the  α - phase and γ-phase. Now, let us look at the two phrases:

α - phase

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γ-phase

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4 years ago
What will happen if water at 50 degree is cooled to -10 degree?
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8 0
4 years ago
Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2
quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure

V_1 = initial volume of gas

V_2 = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 25^oC=273+25=298K

V_1 = initial volume of gas

V_2 = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(2.5atm)\times (10-1)L

w=-22.5L.atm=-22.5\times 101.3J=-2279.25J

(2) A free isothermal expansion from 1 L to 100 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})

w=-11409.6J

(3) A reversible isothermal expansion from 0.5 L to 4 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})

w=-5151.97J

(4) A reversible isothermal expansion from 0.5 L to 40 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})

w=-10856.8J

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(0.5atm)\times (100-1)L

w=-49.5L.atm=-49.5\times 101.3J=-5014.35J

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0
3 years ago
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