The mass of the electron is..

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

3 years ago

. A reaction in which A, B, and C react to form products is first order in A, second order in B, and zero order in C. a) Write a

AleksAgata [21]

Answer:

a) r = k × [A] × [B]²

b) 3

Explanation:

Let's consider the following generic reaction

A + B + C ⇒ Products

The generic rate law is:

r = k × [A]ᵃ × [B]ᵇ × [C]ⁿ

where

r: reaction rate

k: rate constant

a, b, n: reaction orders

This reaction is first order in A, second order in B, and zero order in C. The rate law is:

r = k × [A]¹ × [B]² × [C]⁰

r = k × [A] × [B]²

The overall order of the reaction is the sum of the individual reaction orders.

1 + 2 + 0 = 3

4 0

3 years ago

What would be the major product obtained from hydroboration–oxidation of the following alkenes?

zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

8 0

3 years ago

For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Ag + is reduced to Ag . Complete and balance the equation for t

Sunny_sXe [5.5K]

Answer:

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺ → Ag

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ → Ag (1)

Now, we need to balance the half-reaction of Cr:

Cr → CrO₄²⁻

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:

Cr + OH⁻ → CrO₄²⁻ + 6e⁻

The hydrogen atoms will be balanced using H₂O molecules:

Cr + OH⁻ → CrO₄²⁻ + 6e⁻ + H₂O

The balanced equation is:

Cr + 8OH⁻ → CrO₄²⁻ + 6e⁻ + 4H₂O (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ → Ag)

Cr + 8OH⁻ → CrO₄²⁻ + 6e⁻ + 4H₂O

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

7 0

4 years ago

If a 0.320 mM solution of MnO41- has an absorbance of 0.480 at 525 nm in a 1.000 cm cell. What is the concentration of a MnO41-

larisa [96]

Answer:

Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.

Explanation:

Now A = el, el=const

Then,

$A2 / A1 = C2/ C1\\\\0.49/ 0.48 = C2 / 0.32\\\\C2 = 0.3266$

3 0

3 years ago