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marta [7]
3 years ago
12

Compare to the sample of helium at STP, the same sample of helium at a higher temperature and a lower pressure.

Chemistry
1 answer:
Molodets [167]3 years ago
4 0

Here we have to compare the state of helium gas at STP and high temperature and low pressure.

At STP (standard condition of temperature and pressure) i.e. 273K temperature and 1 bar pressure. At STP helium gas will behave as a real gas.

At higher temperature and low pressure Helium will behave as an ideal gas.

The ideal gas conditions are developed on taking into account two factors: (i) the gas molecules are point of mass and having no volume. (ii) there is no existence of force of attraction between the molecules.

The deviation from ideal gas to the real gas depends upon the van der waals' interaction between the gas molecules. Now in low pressure and high temperature, we can ignore the volume and also the inter-molecular force of attraction. Thus the gas sample can behaved as ideal gas.

But at elevated pressure and low temperature i.e. STP the assumptions are not valid and it will behave as real gas.    

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Imagine that in another universe, atoms and elements are identical to ours, except that atoms with six valence electrons have pa
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Answer:

<em>a. Aluminium</em>

<em>b.</em> <em>Florine</em>

<em>c. oxygen</em>

Explanation:

NB: This answer is based on the assumption that, just like our universe, in this alternate universe, the innermost shell of elements is filled first, and is stable with just two electrons.

A reactive metal will be a metal element that has one electron less than that needed to achieve valency. The metal will tend to want to gain electron from a non-metal to achieve a stable outer shell. For aluminium, a metal, with an atomic number of 13; in this alternate universe, <em>the electronic configuration will be 2, 6, 5</em>. The outermost shell will need an additional electron from a non-metal to achieve valency, making it very reactive with a non metal.

A reactive non-metal will have its outermost shell having just one electron, the non-metal will tend to want to lose this electron to a metal to achieve valency with the already filled inner shell. For Florine, a non-metal, with an atomic number of 9; in this alternate universe, <em>the electronic configuration will be 2, 6, 1</em>. Losing the outermost electron, to a metal, frees up the inner complete stable shell, making it very reactive with a metal.

A noble gas is an element, usually in gaseous form, that has a complete outer shell, and hence has no need for gaining or losing extra electron(s). In an alternate universe, Oxygen, will have an<em> electronic configuration of 2, 6. </em>This is stable and hence it does not react with other elements.

8 0
4 years ago
P4 (s) + 5O2 (g) ⟶ P4 O10(s) ΔG° = −2697.0 kJ/mol 2H2 (g) + O2 (g) ⟶ 2H2 O(g) ΔG° = −457.18 kJ/mol 6H2 O(g) + P4 O10(s) ⟶ 4H3 PO
aalyn [17]

Answer:

the standard free energy of formation of phosphoric acid H3 PO4 is -1010 kJ/mol

Explanation:

Knowing that

1) P4 (s) + 5O2 (g) ⟶ P4 O10(s) ΔG° = −2697.0 kJ/mol

2) 2H2 (g) + O2 (g) ⟶ 2H2 O(g) ΔG° = −457.18 kJ/mol

3) 6H2 O(g) + P4 O10(s) ⟶ 4H3 PO4 (l) ΔG° = −428.66 kJ/mol

since

ΔG° reaction = ν * ΔGf °  products - v *ΔGf °  reactives

for reaction 1

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 1 *ΔGf ° P4 O10 - ( 5 *ΔGf ° O2 + 1 *ΔGf ° P4)

ΔG° = ΔGf ° P4 O10 - (5*0 +1*0)

ΔGf ° P4 O10 = ΔG° = −2697.0 kJ/mol

for reaction 2

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 2 *ΔGf ° H20 - ( 1*ΔGf ° O2 + 2 *ΔGf ° H2)

ΔG° = 2* ΔGf ° H20 - (1*0 +2*0)

ΔGf ° H20 = ΔG° /2  = -457.18 kJ/mol/2 = -228.59 kJ/mol

for reaction 3

ΔG° = ∑ν * ΔGf °  products - ∑v *ΔGf °  reactives

ΔG° = 4 *ΔGf ° H3 PO4 - ( 6*ΔGf ° H2O + 1*ΔGf ° P4O10)

−428.66 kJ/mol = 4 *ΔGf ° H3 PO4 - [ 6*(-228.59 kJ/mol) + 1*(−2697.0 kJ/mol)]

−428.66 kJ/mol = 4 *ΔGf ° H3 PO4 + 3611.36 kJ/mol

ΔGf ° H3 PO4 = (−428.66 kJ/mol - 3611.36 kJ/mol)/4 = -1010 kJ/mol

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