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worty [1.4K]
3 years ago
15

If you have 1.005 x 1024 molecules of Ca(OH)2 how many atoms do you have? Moles? Grams?

Chemistry
1 answer:
ira [324]3 years ago
4 0

Answer:

  • 5.025x10²⁴ atoms
  • 1.6686 moles
  • 123.63 g

Explanation:

In one Ca(OH)₂ molecule, there are 5 atoms (1 Ca, 2 O and 2 H). This means that in 1.005x10²⁴ Ca(OH)₂ molecules, there are (5 * 1.005x10²⁴) 5.025x10²⁴ atoms.

To <u>convert molecules into moles</u>, we use <em>Avogadro's number</em>:

  • 1.005x10²⁴ molecules * \frac{1mol}{6.023*10^{23}molecules} = 1.6686 moles

Finally, we <u>convert moles into grams</u> using the <em>molar mass of Ca(OH)₂</em>:

  • 1.6686 moles * 74.093 g/mol = 123.63 g
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Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol  

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

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<h3>What is the bond angle?</h3>

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The bond angle between the atoms of phosphine is 93°. It has one lone pair. The central atom is covered with 4 atoms.

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