How many protons does Royalty, R-13, have?

Unit 1- Unit Assessment (Matter & Density)

konstantin123 [22]

Answer:

<h3>The answer is 11 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 3025 g

volume = 275 mL

We have

$density = \frac{3025}{275} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

What is alchemy?

Naily [24]

Answer: A. An early form of chemistry that people used to try to turn metal

into gold.

Explanation:

Alchemy is indeed an early form of chemistry in which people try to turn metal into gold.

6 0

4 years ago

Which of the slightly soluble salts below will be more soluble in acidic solution than in pure water?

AlexFokin [52]

Answer:

h. both Mg(OH)₂ and CaCO₃

Explanation:

Let's consider the solution of Mg(OH)₂ according to the following equation:

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

In acidic solution, OH⁻ reacts with H⁺ to form H₂O.

OH⁻(aq) + H⁺(aq) ⇄ H₂O(l)

According to Le Chatelier's principle, since [OH⁻] decreases, the solution of Mg(OH)₂(s) shifts toward the right, increasing its solubility.

Let's consider the solution of CaCO₃ according to the following equation:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

In acidic solution, CO₃²⁻ reacts with H⁺ to form HCO₃⁻.

CO₃²⁻(aq) + H⁺(aq) ⇄ HCO₃⁻(aq)

According to Le Chatelier's principle, since [CO₃²⁻] decreases, the solution of CaCO₃(s) shifts toward the right, increasing its solubility.

Let's consider the solution of AgCl according to the following equation:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Cl⁻ does not react with H⁺ because it comes from a strong acid (HCl). Therefore, the solubility of AgCl(s) is not affected by the pH.

6 0

3 years ago

If an object is not accelerating, you can be sure that

lana66690 [7]

Answer:

A

Explanation:

It's not moving so it would be going 0 meters per second

6 0

3 years ago

Lead all chlorate is mixed with hydrolylic acid. Each solution is 0.85 molar. Write balanced, molecular, ionic, and net equation

d1i1m1o1n [39]

Answer:

Here's what I get

Explanation:

Solubility rules

Salts containing halides are generally soluble. Important exceptions to this rule are halides of silver, mercury, and lead(II).
All acetates, chlorates, and perchlorates are soluble

So, PbCl₂ is insoluble, and Pb(ClO₃)₂ is soluble.

1. "Molecular" equation

$\rm Pb(ClO_{3})_{2}(aq) + 2HCl(aq) \longrightarrow \, PbCl_{2}(s) + 2HClO_{3}(aq)$

2. Ionic equation

Convert the soluble salts to their hydrated ions.

HCl and HClO₃ are strong acids. Convert them to their ions.

$\rm Pb^{2+}(aq) + 2ClO_{3}^{-}(aq)+ 2H^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + 2H^{+}(aq) + 2ClO_{3}^{-}(aq)$

3. Net ionic equation

Cancel all ions that appear on both sides of the reaction arrow (in boldface).

$\rm {Pb}^{2+}(aq) + \textbf{2ClO}_{3}^{-}(aq)+ \textbf{2H}^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + \textbf{2H}^{+}(aq) + \textbf{2ClO}_{3}^{-}(aq)$

The net ionic equation is

$\rm {Pb}^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s)$

4. Theoretical yield

We have the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

(i). Gather all the information in one place with molar masses above the formulas and masses below them.

M_r: 278.11

Pb(ClO₃)₂ + 2HCl ⟶ PbCl₂ + 2HClO₃

Volume/mL: 125 95

c/mol·L⁻¹: 0.85 0.85

(ii) Calculate the moles of each reactant

$\text{Moles of Pb(ClO$_{3}$)}_{2} = \text{0.125 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.1062 mol}\\\text{Moles of HCl} = \text{0.095 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.08075 mol}$

(iii) Identify the limiting reactant

Calculate the moles of PbCl₂ we can obtain from each reactant.

From Pb(ClO₃)₂:

The molar ratio of PbCl₂:Pb(ClO₃)₂ is 2:2

Moles of PbCl₂ = 0.1062 × 2/2 =0.1062 mol PbCl₂

From HCl :

The molar ratio of PbCl₂:HCl is 1 mol PbCl₂:2 mol HCl.

Moles of PbCl₂ = 0.08075 × 1/2 = 0.04038 mol PbCl₂

The limiting reactant is HCl because it gives the smaller amount of PbCl₂.

(iv) Calculate the theoretical yield of PbCl₂.

$\text{Theor. yield of PbCl}_{2} = \text{0.0438 mol} \times \dfrac{\text{278.11 g}}{\text{ 1 mol}} = \textbf{11.2 g}$

5. Calculate the actual yield of PbCl₂

$\text{Actual yield} = \text{11.2 g theor.} \times \dfrac{\text{ 68 g actual}}{\text{100 g theor,}} = \textbf{7.6 g}$

6. Calculate [ClO₃⁻]

Original concentration of Pb(ClO₃)₂ = 0.85 mol·L⁻¹

Original concentration of ClO₃ = 2 × 0.85 = 1.70 mol·L⁻¹

The solution was diluted by the addition of HCl.

Total volume = 125 + 95 =220 mL

c₁V₁ = c₂V₂

1.70 mol·L⁻¹ × 125 mL = c₂ × 220 mL

212.5 mol·L⁻¹ = 200 c₂

c₂ = (212.5 mL)/200 = 1.06 mol·L⁻¹

7. Calculate [Pb²⁺].

Moles of Pb²⁺ originally present = 0.1062 mol

Moles of Pb²⁺removed = 0.04038 mol

Moles of Pb²⁺ remaining = 0.0659 mol

c = 0.0659 mol/0.220 L = 0.299 mol·L⁻¹

8 0

4 years ago