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ankoles [38]
2 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]2 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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A liquid has a volume of 34.6 ml and a mass of 46.0 g. what is the density of the liquid?
Minchanka [31]

The density of a substance can simply be calculated by dividing the mass by the volume:

density = mass / volume

 

Therefore calculating for the density since mass and volume are given:

density = 46.0 g / 34.6 mL

density = 1.33 g / mL

5 0
2 years ago
A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire
melamori03 [73]

Moles of Copper =

1.52×10^{22}atoms×\frac{1 mol }{6.022 * 10^{23}atoms}

= 0.0252 mol Cu

6 0
2 years ago
Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.
Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means  1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × \frac{254 g}{1 mole}= 508 g of Phenobarbital sodium.

1 L ×\frac{1000 mL}{ 1 L} = 1000 mL of solution

Thus, % w/v is:

\frac{508 g}{1000 mL} × 100 = 50,8 % w/v

And radio strengths:

\frac{1000 mL}{508 g}  = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0
2 years ago
Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b
lakkis [162]

Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).

Explanation :

The number of bonding electrons and non-bonding electrons in the structure of XeF_2 is determined by the Lewis-dot structure.

Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.

In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.

Xenon has 8 valence electrons and fluorine has 7 valence electrons.

Total number of valence electrons in XeF_2 = 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we conclude that

The number of electrons used in bonding = 4

The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18

Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).

The Lewis-dot structure of XeF_2 is shown below.

4 0
2 years ago
Read 2 more answers
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
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