a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
[H₃O⁺] = 0,004M
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = 0,085M
ka = 1,88x10⁻⁴
And pka = 3,73
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
pkb = 10,27
I hope it helps!
It is the type of cell used in electroplating
It uses an electric current to make a nonspontaneous reaction go.
All of the above
18. The products formed in the net reaction of the electrolysis of water are ____.
aqueous hydrogen ion and hydroxyl ion
hydrogen gas and oxygen gas
liquid hydrogen and oxygen gas
liquid oxygen and hydrogen gas
19. What occurs in electroplating?
decomposition of a metal layer
decomposition of a salt layer
deposition of a metal layer on a material
deposition of a salt layer on a metal
An electrolytic cell uses an electric current to make a nonspontaneous reaction go and is used in electroplating. The products formed in the electrolysis of water are hydrogen gas and oxygen gas. Electroplating involves the deposition of a metal layer on a material.
An electrolytic cell is a type of cell that uses an electric current to make a nonspontaneous reaction go. This means it can drive a chemical reaction in a direction that does not occur naturally. Electrolytic cells are commonly used in electroplating, where a metal layer is deposited onto a material. Therefore, all of the given options in question 17 are true about an electrolytic cell. For question 18, the products formed in the net reaction of the electrolysis of water are hydrogen gas and oxygen gas. Electrolysis of water involves the decomposition of water into its elemental components. In electroplating, the process involves the deposition of a metal layer on a material. This is achieved by passing an electric current through a solution containing metal ions, causing the metal ions to be reduced and deposit onto the material's surface.
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Answer:
It should be used 2 digits to the right of the decimal point to report the result
Explanation:
When adding or subtracting two decimal numbers, the number of digits to the right of the decimal point in the result is equal to the amount with the least number of decimal places.
In this case we have 4 decimal places in 0.0136 g, 2 decimal places in 2.70 × 10-4 g and 2 decimal places in 4.21 × 10-3 g, so, the least number of decimal places is 2 and that should be the number of digits to the right of the decimal point in the result.
Then the result of the sum should be 9.66 × 10-3 g
I got 2.......................
Answer: 1,88×10²³ atoms of Br in 25g of Br
Explanation:
25g Br
1 mol of Br =79,9g Br
Number of Avogadro: 6,022×10²³ = 1 mol
25gBr× 1molBr/79,9gBr =
6,022×10²³ atoms of Br/1 mol Br =
1,88×10²³ atoms of Bromine