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Yakvenalex [24]
3 years ago
14

The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 17

75. In this experiment Lavoisier found that mercury(II) oxide, when heated, decomposed into liquid mercury and an invisible and previously unknown substance: oxygen gas.
Required:
a. Write a balanced chemical equation, including physical state symbols, decomposition of solild mercurytl) oxide (HgO) into liquid mercury and gaseous dioxygen.
b. Suppose 73.0 mL. of dioxygen gas are produced by this reaction, at a temperature of 130.0 °C and pressure of exactly 1 atm. Calculate the mass of mercury() oxide that must have reacted.
Chemistry
1 answer:
yawa3891 [41]3 years ago
5 0

Answer:

a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

b. 0.957 g

Explanation:

Step 1: Write the balanced equation

2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

Step 2: Convert 130.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15

K = 130.0°C + 273.15

K = 403.2 K

Step 3: Calculate the moles of O₂

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K

n = 2.21 × 10⁻³ mol

Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂

The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.

Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO

The molar mass of HgO is 216.59 g/mol.

4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g

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<h3>Stoichiometric calculation</h3>

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Read 2 more answers
A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small
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Answer:

a)   1.12 MeV / nucleon

b)   5.62 MeV / nucleon

c)  8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

<u>a) ²H = 1 neutron , 1 proton = 2 nucleons </u>

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm  = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3  * 931.5 ) MeV / 2 nucleons

                                                  = 1.12 MeV / nucleon

<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 )  = 7.0582 u  

Δm  = ( 7.0582 u  - 7.0160 u  ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

                                                  = 5.62 MeV / nucleon

<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

                    = 62.5142 u

Δm = 0.5859 u  

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

                                                  = 8.80 MeV / nucleon

<u>D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons </u>

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

                    = 110.9138 u

Δm  = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

                                                  = 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

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3 years ago
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