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ankoles [38]
3 years ago
8

In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for elect

rons. II. the oxidized form has a lower affinity for electrons. III. the reduced form has a higher affinity for electrons. IV. the greater the tendency for the oxidized form to accept electrons.
Chemistry
1 answer:
qwelly [4]3 years ago
4 0

Answer:

The 1st and 4th options are correct

I.the oxidized form has a higher affinity for electrons

IV. the greater the tendency for the oxidized form to accept electrons

Explanation:

Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.

Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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3 years ago
Write application of isotopes .
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Answer:

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6 0
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A solution of KCIO3 is prepared using 75 grams of the solute in enough water to make 0.250 liters of solution. The gram-formula
charle [14.2K]
Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
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3 years ago
Which statement is correct regarding the reaction ?
larisa86 [58]

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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