All categories

3 years ago

In a redox reaction, reduction is defined by the:

Chemistry

1 answer:

jonny [76]3 years ago

5 0

Answer:

C. Gain in electron(s) resulting in a decrease of oxidation number.

Explanation:

Redox reactions are reactions involving transfer of of electron between two species (reduction specie) and (oxidation species) and change resulting in change in oxidation number.

Reduction in terms of redox reaction is the specie that accepts electron(s) and gets "reduced" since its oxidation state has been reduced.

For example

Cl + e- → Cl⁻

The above reaction is an example of reduction reaction taking place in a redox reaction. We can see that Chlorine oxidation state was changed from (0) to (-1) state.

You might be interested in

For a concentration technique, Stotal is given as 19.31 ± 0.035, Smb is 0.22 ± 0.008, and kA is 0.154 ± 0.007 ppm–1, where Stota

professor190 [17]

Answer:

Explanation:

8 0

2 years ago

Natural gas and petroleum are mostly alkanes t/f

dolphi86 [110]

Been a year since i was in chem but my money is on true

5 0

3 years ago

Read 2 more answers

What is the summary at the end of an experiment that explains the results

jasenka [17]

Answer:

The conclusion

Explanation:

6 0

3 years ago

En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son

valkas [14]

Answer:

Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

Explanation:

Paso 1: Escribir la ecuación balanceada

BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0

Paso 2: Calcular el cociente de reacción (Qp)

Qp = pBrF × pF₂ / pBrF₃

Qp = 1,50 × 2,00 / 0,0150 = 200

Paso 3: Sacar una conclusión

Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

7 0

3 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

8 months ago