Answer:
Yes it is hard. For some of us. I already gave up, but now hes coming back. Id say sometimes its useless to get his attention if he never even pays attention to you anyways. It sucks because its so hard to get then to notice you for once.
Explanation:
Answer: The correct answer is: [B]:
_________________________________________________
" organic acid and amines " .
_________________________________________________
<u>Note</u>: Choice B: "organic acid and amines" ;
is the only answer choice that contains "amines" (hint: <u> amin</u><u>o acid</u> / <u>amin</u><u>e)</u> ; which are "proteins" .
As such; Choice "B" is the <u><em>only</em></u> correct answer choice.
_____________________________________________________
Hope this helps!
Best wishes to you!
_____________________________________________________
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
Answer: you can use a strainer
Explanation:
