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Tasya [4]
3 years ago
15

Which of these belongs in the outermost shell (energy level) of an atom?

Chemistry
1 answer:
hjlf3 years ago
6 0

Answer:

C)Electrons

Explanation:

Hope this helped!!!

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They are the basic components of life or the basic building blocks of life
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An alien civilization has different names and symbols for six elements. Place each element in the correct location in the period
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An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at
Yakvenalex [24]

Answer:

  • <u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

  • <em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

  • 2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

       2molHgO/1molO_2

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

          pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol

<u>5. Calculate the number of moles of HgO</u>

         \dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO

<u>6. Convert to mass</u>

  • mass = # moles × molar mass

  • molar mass of HgO: 216.591g/mol

  • mass = 0.315mol × 216.591g/mol = 68.3g

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3 years ago
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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
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