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2 years ago

What effect does the anion of an ionic compound have on the appearance of the solution?

Chemistry

1 answer:

choli [55]2 years ago

6 0

Answer: B. The anion affects the color of the solution more than the intensity of the color.

Explanation:

An ionic bond is gotten when an electron is transferred from a metal atom to a non-metal one. It should be noted that the ionic bonds simply has an anion and a cation.

An anion is formed when a valence election is gained by a non metal while a cation is formed when the metal ion misplaces a valence electron.

The effect of the anion of an ionic compound on the appearance of the solution is that the anion affects the color of the solution more than the intensity of the color.

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20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

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As ice melts from a solid to a liquid, A) entropy increase

Dovator [93]

That answer is false

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Molecule: Br2 and Br2. <br> Is it polar or nonpolar?

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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

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