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>
<u><em /></u>
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>
<em></em>
I hope it helps!
Answer:
<h2>15 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

But from the question
volume = final volume of water - initial volume of water
volume = 165 - 150 = 15 mL
We have

We have the final answer as
<h3>15 g/mL</h3>
Hope this helps you
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
There is an excess of oxygen, the two main products should be carbon dioxide gas and water vapor. If the oxygen supply is limited, the fuel will undergo incomplete combustion and produce carbon monoxide, water, and sometimes carbon.
Hope this helped! :)
Answer:
Iconic bonds don't burn easily
Explanation:
Covalent bonds are non metals. Covalent bond (sharing), low temp, low temp, burn easily, poor, polar covalent is good and non-polar covalent is bad.
Ionic - metals and nonmetals, ionic bond is when electrons are gained or lost, high temp, high temp, doesn't burn easily, good, good