PLEASE HELP !! ILL GIVE BRAINLIEST !! 100 POINTS IF YOUR GOOD AT CHEMISTRY AND BIOLOGY .

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago

Five solutions on how to prevent air pollution

I am Lyosha [343]

Answer:

Reduce number of trips

Avoid burning leaves

Avoid using garden equipments

Reduce the use of wood stove

Avoid gas-powered lawn.

Explanation:

7 0

3 years ago

Identify each of the following solids as molecular, ionic, or atomic.

Kisachek [45]

Answer :

(A) Br₂ (s) : molecular solids

(B) AgCl (s) : ionic solids

(C) S (s) : atomic solids

(D) CH₄ (s) : molecular solids

Explanation :

Molecular solids : It is defined as the solids in which they are held together by covalent forces, dipole interactions as attractive forces etc.

Ionic solids : It is defined as the solids in which the atoms composed with oppositely charged ions.

Atomic solids : It is defined as the solids in which the molecules are held together by covalent forces and also includes pure substance.

(A) Br₂ (s)

It is molecular solids because they are held together by covalent forces.

(B) AgCl (s)

It is ionic solids because in this atoms composed with oppositely charged ions.

(C) S (s)

It is atomic solids because it is a pure substance.

(D) CH₄ (s)

It is molecular solids because they are held together by covalent forces.

7 0

3 years ago

If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically

Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the equation

NaOH = $\frac{842}{39.9}$

= 21.1 MOLES OF NaOH

Al = $\frac{750}{26.9}$

= 27.8 moles

from the equation

from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

$\frac{1}{3}$ = $\frac{x}{21.1}$

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

= 548.55 grams

theoretical yield from the given data is 548.55 grams

3 0

3 years ago

A 45.0 g sample of a metal at 85.6 Â°C is placed in 150.0 g of water at 24.6 Â°C. The final temperature of the system is 28.3 Â°

earnstyle [38]

Answer:

904.014 j/kgk

Explanation:

Mass of metal = 45g

Temperature of metal = 85.6°

Mass of water = 150

Temperature of water = 24.6

Final temperature of system = 28.3

Heat lost by metal = Heat gained by water

m1 * c1 * dt = m2 * c2 * dt

Q = quantity of heat

Q = m*c*dt

dt = change in temperature

dt of water = 28.3 - 24.6 = 3.7

dt of metal = 85.6 - 28.3 = 57.3

Specific heat capacity of water, c = 4200

(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7

2.5785c1 = 2331

c1 = 2331 / 2.5785

= 904.01396

= 904.014 j/kgk

3 0

3 years ago