How do I do this organ journal ?

Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter

maksim [4K]

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(A) The given balanced reaction is,

$H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

$S_8(s)+8O_2(g)\rightarrow 8SO_2(g)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

$3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)$

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

$Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

$PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

6 0

3 years ago

The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2

pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0

3 years ago

10 points please please help

Inga [223]

Your answer would be respiration. This would be your answer because, plants take in carbon dioxide, and animals, including humans, breath oxygen. I hope this had helped you out.

4 0

3 years ago

A 35.66g sample of copper is heated using 600j of energy. if the original temperature of the copper is 85C what is its final tem

ira [324]

This problem is providing the mass, energy, initial temperature and specific heat of a sample of copper that is required to calculate the final temperature.

Thus, we recall the general heat equation:

$Q=mC(T_f-T_i)\\$

Which has to be solved for the final temperature, $T_f$ as follows:

$T_f=T_i+\frac{Q}{mC}$

Finally, we plug in the numbers to obtain:

$T_f=85\°C+\frac{600J}{35.66g*0.38\frac{J}{g\°C} } \\\\T_f=129.3\°C$

However, this result is not given in the choices.

Learn more:

brainly.com/question/14383794

8 0

2 years ago

A hydrocarbon of pentane has two isomeric forms—n-pentane and iso-pentane. The molecular formula of the n-pentane is C5H12. Pred

Monica [59]

An iso pentane is an isomeric form of n-pentane. Isomers have same molecular formula but differ in the atomic arragement projected in 3D and the arrangement of atoms primarily resulting to a different set of properties. In this case, isopentane also has a molecular formula of an alkane <span>C5H12</span>

5 0

3 years ago