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Pepsi [2]
3 years ago
9

Give 2 examples of compounds.

Chemistry
2 answers:
Andreas93 [3]3 years ago
6 0

Answer:

the answer is salt and water.

Explanation:

LiRa [457]3 years ago
3 0

Answer:

salt and water

Explanation:

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How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?
Bad White [126]

The balanced chemical reaction is :

O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na, n = \dfrac{14.6}{23} = 0.635 \  mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

n = \dfrac{0.635}{4}\  mole

So, amount of oxygen required is :

m = \dfrac{0.635 \times 32}{4}\  gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0
3 years ago
How is radiation different from conduction?
steposvetlana [31]

Answer:

<u>Radiation is the transfer of energy by waves, and conduction is the transfer of heat through contact with air.</u>

Explanation:

Conduction is the transfer of thermal energy through direct contact.  Radiation is the transfer of thermal energy through thermal emission.

7 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

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