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german
4 years ago
15

Two different compounds, each containing only phosphorus and chlorine, were decomposed into their constituent elements. One prod

uced 0.2912 g P for every gram of Cl; the other produced 0.1747 g P for every gram of Cl. Show that these results are consistent with the law of multiple proportions.
Chemistry
1 answer:
Katarina [22]4 years ago
5 0

Answer:

Dalton's law is fulfilled, since the elements in each compound of the example are related to each other in different proportions forming two different compounds;  so for case a) and b) you have 15.54% and 9.323% P respectively for each compounds,  that are equivalent to 0.2912 g P for every g Cl and the other produced 0,1747 g P for every g Cl.

Explanation:

Dalton's Law ( law of multiple proportions):

When two or more elements combine to give more than one compound, a variable mass of one of them joins a fixed mass of the other, and the first is related to canonical and indistinct numbers.

  • P + Cl ↔ PCl

a) for the compounds A:

⇒ 0.2912 g P / gCl * ( 35.45 g CL / mol CL) * ( mol Cl / mol PCl ) * ( mol PCl / 66.427 g PCl ) = 0.1554 g P / g PCl

⇒ % P = 0.1554 * 100 = 15.54 %

b) for the compunds B:

⇒ 0.1747 g P / g Cl * ( 35.45 g Cl / mol Cl ) * ( mol Cl / mol PCl ) * ( mol PCl /  66.427 g PCl ) = 0.09323 g P / g PCl

⇒ % P = 9.323 %

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Answer:

113.69°k

Explanation:

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3 years ago
The kb of hypochlorous acid is 3. 0×10^–8 at 26. 0 C. What is the percent of ionization of hypochlorous?
astraxan [27]

The acid dissociation constant (Ka) defines the difference between a weak and a strong acid. The % ionization of hypochlorous acid is 0.14%.

<h3>What is the acid dissociation constant?</h3>

The acid dissociation constant is used to define the ionization constant of an acidic substance. It gives the quantitative measurement of the strength.

The ICE table is attached to the image below.

The acid dissociation constant (Ka) for the reaction is,

Ka = [H⁺][ClO⁻] ÷ [HClO]

= a² ÷ (0.015 - a)

= 3.0 x 10⁻⁸

Now, a² + 3.0 x 10⁻⁸ a - 4.5 × 10⁻¹⁰ = 0

So, a = 2.210 × 10⁻⁵

Solving further,

[H+] = a = 2.210 × 10⁻⁵ M

The percent ionization is calculated as,

[H+] ÷ [HClO] × 100

= 2.210 × 10⁻⁵ M ÷ 0.015 × 100

= 0.14 %

Therefore, 0.14 % is the percentage of hypochlorous ionization.

Learn more about acid dissociation constant here:

brainly.com/question/22668939

#SPJ4

Your question is incomplete, but most probably your full question was, The ka of hypochlorous acid (HClO) is 3.0 x 10⁻⁸ at 25.0°C. What is the % of ionization of hypochlorous acid in a 0.015 aqueous solution of HClO at 25.0C?

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Respuesta:

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Consideremos la ecuación no balanceada que ocurre cuando cloruro férrico acuoso reacciona con carbonato de sodio sólido para formar carbonato férrico sólido y cloruro de sodio acuoso. Esta es una reacción de doble desplazamiento.

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Vamos a usar el método de tanteo. Empezaremos balanceando los átomos de C, multiplicando Na₂CO₃ por 3.

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Luego, balancearemos los átomos de Fe, multiplicando FeCl₃ por 2.

2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)

Finalmente, obtendremos la ecuación balanceada, multiplicando NaCl por 6.

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