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Helga [31]
3 years ago
6

What happens in a reduction half-reaction?

Chemistry
2 answers:
Marta_Voda [28]3 years ago
3 0
The answer is elements gain electrons. Oxidation reduction is elements lose electrons. And oxygen is added/lost can be a type of oxidation/reduction reaction.
Oksanka [162]3 years ago
3 0

Answer: Option (a) is the correct answer.

Explanation:

A reduction-half reaction is defined as the reaction in which there occurs gain of one or more number of electrons by a specie or element.

Or, in a reduction-half reaction there occurs loss of oxygen if it is present in a reaction.

But in a reduction-half reaction loss of electrons will always take place.

For example, CuO + Mg \rightarrow Cu + MgO

Reduction-half reaction: Cu^{2+} + 2e^{-} \rightarrow Cu

So here, oxidation state of copper is changing from +2 to 0. Also, loss of oxygen is occurring in the reaction equation.

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

Thus, we can conclude that in a reduction half-reaction elements gain electrons.

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A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q_1=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

6 0
3 years ago
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Complete the hybridization and bonding scheme for xef4.
Vikentia [17]

<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>

total ve = 8 + 28 = 36 ve

 

<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>

 

<span>28 - 4(6) = 4 
(We assign the remaining electrons to F atoms)</span>

 

<span>4 - 2(2) = 0 
(Therefore 4 electrons left => we have 2 lone pairs)</span>

 

The steric number = No. of σ bonds + #lone pairs

= 4 σ bonds + 2 lone pairs = 6 => d²sp³ (6 hybrid orbitals)

 

<span>4 bonds + 2 lone pairs => square planar</span>

6 0
3 years ago
Which value of gas constant (R)use and when ?<br>in ideal gas equation. ​
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Answer:

R is universal gas constant

It is has fixed value which would depend upon the units in which P,V and T expressed in the equation PV=nRT

R=

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4 0
2 years ago
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ANTONII [103]

Answer:

True

Explanation:

When a base is dissolved in water, the balance between hydrogen ions and hydroxide ions shifts the opposite way. Because the base "soaks up" hydrogen ions, the result is a solution with more hydroxide ions than hydrogen ions. This kind of solution is alkaline.

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Help me please it’s all due soon
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Omg we did this but i forgot how to do it oh and i failed the test so um try to type the same problem on google that's how i got all the right answer on my homework. Watch this video tells you the answer https://youtu.be/S8GjkD296gg

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