# Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?

The correct answer is 5.6 × 10⁻²³ M.

Explanation:

As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.

Let the solubility of Hg₂Br₂ is S mol per liter.

Therefore,

KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)

Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)

Net [Br-] = (2S + 1) M

Ksp = S (2S + 1)²

Ksp = S (4S² + 1 + 4S)

Ksp = 4S³ + S + 4S²

As the solubility is extremely less, therefore, we can ignore S² and S³. Now,

Ksp = S = 5.6 × 10⁻²³ M

Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.

## Related Questions

When a polar bond is formed between 2 atoms which atom receives a partial positive charge

The less electronegative atom

Explanation:

Water contains polar bonds. Water is made up of oxygen and hydrogen. Oxygen is more electronegative then hydrogen so it is partial negative (because electron spend more time around oxygen in the polar bond), whereas, hydrogen has a  partial positive charge as it is less electronegative (electrons spend less time around hydrogen in this polar bond).

I'm not an expert at this, but I assume its mercury.

A piece of charcoal used for cooking is found at the remains of an ancient campsite. a 0.94 kg sample of carbon from the wood has an activity of 1580 decays per minute. find the age of the charcoal. living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14c is 5730 y. answer

Mass of sample of charcoal = 0.94 kg = 0.00094

∴, activity = decay rate / mass = 1580/0.00094
= 1.681 X 10^6 decays per min per gram

Using the half-life formula, we have:
activity of sample / activity of modern carbon = (1/2)^(age / half-life)
∴, Age = half-life x log (base 2) (modern activity / coal activity)
= 5730 x log(base 2)(1.681X10^6/ 15)
= 96115 years.

Answer: Age of the charcoal = 96115 years

Using the radiocarbon dating technique and applying the decay formula, it is calculated that the age of the charcoal from the an ancient campsite is approximately 9,500 years.

### Explanation:

The age of the charcoal can be found using the technique of radiocarbon dating, which capitalizes on the process of radioactive decay. The isotope carbon-14 (¹4C) is used in this method as it has a known half-life of 5730 years. The number of decays per minute per gram of carbon in a live organism is known as its activity.

Initially, the activity was given as 15 decays per minute per gram. The present activity of the carbon in the charcoal is provided at 1580 decays per minute for a 0.94 kg or 940 gram sample. Thus, the current activity per gram is 1580/940 equals approximately 1.68 decays per minute per gram.

Given that the half-life of ¹4C is 5730 years, we can apply the formula for calculating the time passed using the rate of decay, which is given as T = (t1/2 / ln(2)) * ln(N0/N), where 'ln' is the natural logarithm, 'N0' is the initial quantity (15 decays/minute per gram), 'N' is the remaining quantity (1.68 decays/minute per gram).

Plugging in the given values, we get T = (5730 / ln(2)) * ln(15/1.68), which gives us approximately 9,500 years. Therefore, the age of the charcoal is around 9,500 years.

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What mass of nitrogen gas is required to react completely with 2.79 g of hydrogen gas to produce ammonia?

the balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of H₂ to N₂ is 3:1
number of H₂ moles reacted - 2.79 g / 2 g/mol  = 1.40 mol
if 3 mol of H₂ reacts with 1 mol of N₂
then 1.40 mol of H₂ reacts with  - 1.40/3 = 0.467 mol of N₂
mass of N₂ required - 0.467 mol x 28 g/mol = 13.1 g
mass of N₂ formed is 13.1 g

What is the mass of 0.73 moles of AgNO3?

124 g (3 sig figs)

or

124.011 g (6 sig figs

Explanation:

Step 1: Calculate g/mol for AgNO₃

Ag - 107.868 g/mol

N - 14.01 g/mol

O - 16.00 g/mol

107.868 + 14.01 + 16.00(3) = 169.878 g/mol

Step 2: Multiply 0.73 moles by molar mass

0.73 mol (169.979 g/mol)

124 grams of AgNO₃

What is the molarity of a 27.0% (v/v) aqueous ethanol solution? the density of ethanol (c2h6o, molar mass 46.07 g/mol) is 0.790 g/ml?

Explanation:

Hello,

In this case, with the given by-volume percentage and considering the molarity as:

We assume the solution having 100 mL of volume in total, thus, the volume of ethanol is 27.0 mL, therefore, the moles:

Moreover, the volume of the solution in liters is:

Finally, the molarity is:

Best regards.

If the density of ethanol (c2h6o, molar mass 46.07 g/mol) is 0.790 g/ml, the molarity of the 27.0% (v/v) aqueous ethanol solution is 17.14 M.

Calculate the moles of ethanol contained in the solution, then divide that number by the volume of the solution in litres to determine the molarity of the ethanol solution.

To start, we must ascertain how much ethanol is included in each 100 millilitres of the solution. Given that ethanol has a density of 0.79 g/ml, the amount of ethanol in 100 millilitres is as follows:

Mass of ethanol = density × volume

Mass of ethanol = 0.790 g/ml × 100 ml = 79 g

Now,

Moles of ethanol = mass / molar mass

Moles of ethanol = 79 g / 46.07 g/mol = 1.714 mol

So,

Volume of solution = 100 ml / 1000 ml/L = 0.1 L

We know that:

Molarity = moles of solute / volume of solution

Molarity = 1.714 mol / 0.1 L = 17.14 M

Thus, the molarity is 17.14 M.

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