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Sauron [17]
3 years ago
11

A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in different proportions. Which of the followi

ng pieces of information could be used to determine which mixture has the higher proportion of KCl(s) ?
Chemistry
2 answers:
8090 [49]3 years ago
6 0

Answer:

The information that can be used to determine which mixture has the higher proportion of KCl IS INFORMATION ABOUT THE MASS OF CHLORINE IN EACH MIXTURE, THIS INFORMATION CAN BE OBTAINED BY USING THE LAW OF DEFINITE PROPORTION.

Explanation:

The law of definite proportion states that the chemical composition by mass of a chemical compound is always constant. For instance, a chemical compound that is made up of two elements will always contain the same proportions of the constituent elements regardless of the quantity of chemical that was used.

Using the law of definite proportion, we can determine the proportion of sodium and chlorine in NaCl and the proportion of potassium and chlorine in KCl if the mass of chlorine that was used is known. Based on the results obtained, one can easily determine the mixtures that has higher proportion of KCl.

LenaWriter [7]3 years ago
5 0

Answer:

The mass of  Cl  in each mixture

Explanation:

With the amount of Chlorine in the mixture, you can figure out the amount of Potassium and Souium to see which is more.

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Will BeCO3 Hydrolyze in aqueous medium?. If so how?... very urgent..​
irinina [24]

Answer:

BeCO3 will hydrolyze in an aqueous medium, yielding a basic solution.

Explanation:

In accord with the rules of salt hydrolysis, the cation Be^{2+} can hydrolyze. The anion CO_{3} ^{2-} can also hydrolyze, given that the resulting conjugate acid is weak.

Given that both ions hydrolyze, the Ka (acid ionization constant) and Kb (Base ionization constant) of each reactant (or product [that is, the conjugate acid and base for each ion]) must be compared.

Using the web (hopefully some sort of reference table has been provided to you):

Ka (for Be^2+) = 3 * 10^{-7}

Kb (for CO3^2-) = \frac{Kw (water)}{Ka (HCO3)}

= \frac{1*10^{-14}}{4.7*10^{-11}}

= 2.1 * 10^{-4}

The Kb > Ka, so the solution will be basic.

Hope this helps! My apologies if this answer is incorrect, I have not done this type of problem in a while.

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Answer:

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Explanation:

ΔE = hc/λ = (6.63 x 10⁻³⁴j·s)(2.99792 x 10⁸m/s)/(1.73 x 10⁻⁷m) = 1.15 x 10⁻¹⁸j/e⁻ x 6.023 x 10²³ e⁻/mole = 691,999 j/mole e⁻ = 692 Kj/mole e⁻ x 6.24 x 10²¹ eV/KJ = 4.32 x 10²⁴ eV/mole e⁻

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Answer:

pH = -(log 10^-10)

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