Answer:

Answer:

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.

A solid is hard brittle and electrically nonconducting. it's melt ( the liquid form of the substance) and an aqueous solution containing the substance conduct electricity. classify solid.

1. What kind of intermolecular forces act between a chlorine monofluoride molecule and a hydrogen bromide moleculeNote: If there is more than one type of intermolecular force that acts, be sure to list them all, with a comma between the name of each force.

Question 16 Unsaved Which of the following diagrams represents the correct cycling of gases?A) Photosynthesis O2 Respiration CO2 B) Respiration O2 photosynthesis CO2 C) Photosynthesis H2 Respiration O2 D) Respiration CO2 photosynthesis H2

Predict the products and write balanced net ionic equations for the following reactions. (g) SnCl2 is added to KMnO4 solution (acidic) forming Mn2

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

1. What kind of intermolecular forces act between a chlorine monofluoride molecule and a hydrogen bromide moleculeNote: If there is more than one type of intermolecular force that acts, be sure to list them all, with a comma between the name of each force.

Question 16 Unsaved Which of the following diagrams represents the correct cycling of gases?A) Photosynthesis O2 Respiration CO2 B) Respiration O2 photosynthesis CO2 C) Photosynthesis H2 Respiration O2 D) Respiration CO2 photosynthesis H2

Predict the products and write balanced net ionic equations for the following reactions. (g) SnCl2 is added to KMnO4 solution (acidic) forming Mn2

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

**Answer:**

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).

**Answer:**

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

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

∴, 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.

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.**

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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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

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

**Answer:**

**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

**Answer:**

**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.

For more details regarding **molarity**, visit:

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