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Step2247 [10]
3 years ago
12

What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D

) 2.38 x 10-228 E) 1.16 x 10-26
Chemistry
2 answers:
bekas [8.4K]3 years ago
7 0

Answer: D) 2.38*10^-^2^2 g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})

= 2.38*10^-^2^2 g

So, the correct option is D) 2.38*10^-^2^2 g

Vedmedyk [2.9K]3 years ago
6 0

Answer : The mass in grams of a single formula unit of silver chloride will be 2.38\times 10^{-22}g

Explanation:

Formula units : It is defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is 6.022\times 10^{23}.

Single formula unit of silver chloride means that 1 mole of silver chloride.

We are given :

Number of moles of silver chloride = 1 moles

The molar mass of silver chloride = 143.32 g/mole

That means, 1 mole silver chloride contains 143.32 grams of silver chloride.

Now we have to determine the mass in grams of a single formula unit of silver chloride.

As, 6.022\times 10^{23} formula unit has mass = 143.32 g

So, 1 formula unit has mass = \frac{1}{6.022\times 10^{23}}\times 143.32=2.38\times 10^{-22}g

Therefore, the mass in grams of a single formula unit of silver chloride will be 2.38\times 10^{-22}g

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jonny [76]

Answer:

[Ag^+]=2.82x10^{-4}M

Explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M

Now, we can set up the equilibrium expression as shown below:

Ksp=8.51x10^{-17}=(x)(x+2.82x10^{-4})

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

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frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

            KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

  1. Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
  2. The charge on simple ions signifies their oxidation number.
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The oxidation number of K in KClO₂:

                                   K + (-1) + 2(-2) = 0

                                    K-5 = 0

                                    K = +5

The oxidation number of K in KCl:

                                K + (-1) = 0

                                K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

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