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Sonja [21]
3 years ago
14

How many atoms of mercury are present in 3.4 cubic centimeters of liquid mercury? the density of mercury is 13.55 g/cc?

Chemistry
1 answer:
maks197457 [2]3 years ago
8 0

Answer:

             1.37 × 10²³ Atoms of Mercury  

Solution:

Step 1: Calculate Mass of Mercury using following formula,

                               Density  =  Mass ÷ Volume

Solving for Mass,

                               Mass  =  Density × Volume

Putting values,

                               Mass  =  13.55 g.cm⁻³ × 3.4 cm³                ∴ 1 cm³ = 1 cc

                               Mass  =  46.07 g

Step 2: Calculating number of Moles using following formula;

                               Moles  =  Mass ÷ M.mass

Putting values,

                               Moles  =  46.07 g ÷ 200.59 g.mol⁻¹

                               Moles  =  0.229 mol

Step 3: Calculating Number of Atoms using following formula;

                               Number of atoms  =  Moles × 6.022 ×10²³

Putting value of moles,

                               Number of Atoms  =  0.229 mol × 6.022 × 10²³

                              Number of Atoms  =  1.37 × 10²³ Atoms of Hg

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How many grams of water can be produced when 11.7 moles of ethane (C2H6) react with excess oxygen gas?
Elina [12.6K]

Answer: 631.8 g

Explanation:

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

It can be seen from the balanced chemical equation, 2 moles of ethane reacts with 7 moles of Oxygen gas to produce 4 moles of carbon dioxide and 6 moles of water.

Ethane is the limiting reagent as it limits the formation of product.

Thus, if  2 moles of ethane produce 6 moles of water.

11.7 moles moles of ethane produce=\frac{6}{2}\times 11.7=35.1 molesof water.

Mass of water= no of moles\times Molar mass

Mass of water= 35.1\times 18g/mol= 631.8 g

4 0
3 years ago
Read 2 more answers
What is the mass if 1.72 moles of sodium nitrate
alexgriva [62]

Answer:

146 g

Explanation:

Step 1. Calculate the <em>molar mass</em> of NaNO₃

Na =                    22.99

  N =                     14.01

3O = 3 × 16.00 = 48.00

               Total = 85.00 g/mol

Step 2. Calculate the <em>mass</em>  of NaNO₃

Mass of NaNO₃ = 1.72 × 85.00/1

Mass of NaNO₃ = 146 g

8 0
3 years ago
Read 2 more answers
Dilute 0.407M &amp; 2.56L solution to 7.005L. CF?
Alenkinab [10]

Answer:

C₂ = 0.149 M

Explanation:

Given data:

Initial concentration = 0.407 M

Initial volume = 2.56 L

Final volume = 7.005 L

Final concentration = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Initial concentration

V₁ = Initial volume

C₂ = Final concentration

V₂ =Final volume

Now we will put the values.

0.407 M × 2.56 L =  C₂ × 7.005 L

1.042 =  C₂ × 7.005 L

C₂ = 1.042 M.L / 7.005 L

C₂ = 0.149 M

5 0
3 years ago
Which best describes the oxidizing agent in this reaction? Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq) Bromine (Br) is the oxidizing age
steposvetlana [31]
Answer:
            <span>Chlorine (Cl) is the oxidizing agent because it gains an electron.

Explanation:
                   Reaction is as follow,

</span><span>                     Cl</span>₂<span> (aq)  +  2 Br</span>⁻<span> (aq)    </span>→   <span> 2Cl(aq)  +  Br</span>₂ <span>(aq)

Oxidation Reaction:

                                    2 Br</span>⁻     →     Br₂  +  2 e⁻

Two atoms of Br⁻ (Bromide) looses two electrons to form Br₂ molecule. Hence it is oxidized and is acting as reducing agent.

Reduction Reaction:

                                   Cl₂  +  2 e⁻    →     2 Cl⁻

One molecule of Cl₂ gains two electrons to form two chloride ions (Cl⁻). Therefore, it is reduced and has oxidized Br⁻, Hence, acting as a oxidizing agent.
6 0
3 years ago
Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+ (aq) + 2I− (aq) → PbI2 (s) Lead iod
Angelina_Jolie [31]

Answer:

338.00 mL

Explanation:

The lead ions come from the salt Pb(NO₃)₂ and the iodide from the acid HI, so the balanced reaction is:

Pb(NO₃)₂(aq) + 2HI(aq) → PbI₂(s) + 2HNO₃(aq)

So, the stoichiometry is 1 mol of Pb(NO₃)₂ to 2 moles of HI, then:

1 mol of Pb(NO₃)------------------------------------ 2 moles of HI

0.600 mol of Pb(NO₃)₂--------------------------- x

By a simple direct three rule:

x = 1.200 mol of HI

The acid has concentration equal to 3.550 mol/L, the volume (V) is the number of moles divided by the molar concentration:

V = 1.200/3.550 = 0.338 L

V = 338.00 mL

6 0
3 years ago
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