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B. The following reaction takes place in a basic solution. (7 points)

Natali5045456 [20]

Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

$MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)$

To balance the given redox reaction in basic medium, there are few steps to be followed:

Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$

Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$ ( × 3)

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$ ( × 2)

The half-reactions now become:

Oxidation half-reaction: $3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-$

Reduction half-reaction: $2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-$

Add the equations and simplify to get a balanced equation

Overall redox reaction: $3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-$

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0

2 years ago

For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of

brilliants [131]

Answer:

–0.360 M/s

Explanation:

7 0

3 years ago

A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and

AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr

Poctane = 5.12 torr

b)Ptotal vapor = 29.42 torr

c) 81 % heptane

19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

Xa = mole fraction of A

Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

Xa = Pa/Pt where Xa = mol fraction of in the vapor

Pa = partial pressure of A as calculated above

Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)

a)

MW heptane = 100.21 g/mol

MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane and then we can calculate the masses:

0.82 mol x 100.21 g/mol = 82.2 g

0.17 mol x 114.23 g/mol = 19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol) + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then having a mol fraction of 0.83 means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x 114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8 vs 10.9 torr).

4 0

3 years ago

Is sugar a compound?

kvv77 [185]

Sugar is a compound.
C12H22O11

3 0

3 years ago

238,2U --> 234Th + 90

stiv31 [10]

Answer:

$\\_{2}^{4} \alpha$

Explanation:

In this question, we wish to find the missing nuclei for the equation:

$\\_{92}^{238} U\rightarrow _{90}^{234} Th + _{Z}^{A} X$

In order to find the missing species, we need to use the charge and mass balance law. That is, the mass should be conserved: the total mass on the left-hand side with respect to the arrow should be equal to the total mass on the right-hand side with respect to the arrow:

$238 = 234 + A$