Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
Answer:
4. +117,1 kJ/mol
Explanation:
ΔG of a reaction is:
ΔGr = ΔHr - TΔSr <em>(1)</em>
For the reaction:
2 HgO(s) → 2 Hg(l) + O₂(g)
ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)
As ΔHf of Hg(l) and ΔHf O₂(g) are 0:
ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>
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In the same way ΔSr is:
ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)
ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol
ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>
Thus, ΔGr at 298K is:
ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol
ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>
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I hope it helps!
2 valence electrons
Explanation:
Most transition metals have 2 valence electrons. Valence electrons are the sum total of all the electrons in the highest energy level (principal quantum number n). Most transition metals have an electron configuration that is ns2(n−1)d , so those ns2 electrons are the valence electrons.
