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:
Explanation:
C₆H₁₂O₆ + 60₂ = 6H₂0 + 6CO₂.
1 mole 6 mole
molecular weight of C₆H₁₂O₆ = 180 g
molecular weight of oxygen = 32 g
1 gram of glucose = 1 / 180 = 5.55 x 10⁻³ moles
1 gram of oxygen = 1 / 32 = 31.25 x 10⁻³ moles
1 mole of glucose reacts with 6 moles of oxygen
5.55 x 10⁻³ moles of glucose will react with 6 x 5.55 x 10⁻³ moles of oxygen
= 33.30 x 10⁻³ moles of oxygen .
But oxygen available = 31.25 x 10⁻³ moles
So available oxygen is less than required .
Hence oxygen is the limiting reagent .
b ) is the right option .
The answer is B cotton, since you can grow it.
when you get addicted to it and won't stop taking it
You can use physical methods to separate a mixture and also some chemical methods too.
Mixture is just a physical combine.
However a compound is a chemical combine so you must use special methods to separate it