-130KJ is the standard heat of formation of CuO.
Explanation:
The standard heat of formation or enthalpy change can be calculated by using the formula:
standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation
Data given:
Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ
2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ
CuO + Cu ⇒ Cu2O (-11.3 KJ) ( Formation of Cu2O)
When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.
Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ
So standard heat of formation of formation of Cu0 as:
Cu + 1/2 02 ⇒ CuO
putting the values in the equation
ΔHf = ΔH1 + ΔH2 (ΔH1 + ΔH2 enthalapy of reactants)
heat of formation = -11.3 + (-119.35)
= - 130.65kJ
-130.65 KJ is the heat of formation of CuO in the given reaction.
Properties like atomic number, valency etc...
<em>Answer:</em>
<em>An Atom can turn into both solid and liquid form depending on the temperature of its surroundings </em>
<em>Explanation:</em>
<em>Scientists have discovered a new state of physical matter in which atoms can exist as both solid and liquid simultaneously.
Researchers have found, however, that some elements can, when subjected to extreme conditions, take on the properties of both solid and liquid states.</em>
The correct scientific instrument is telescope.
Answer:
a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable
Explanation:
pH = -log[H+] OR pH = -log{H3O+]
and inversely
pOH = -log[OH-]
1. Determine what substance you are working with, (acid/base)
2. Determine whether or not that acid or base is strong or weak.
a. 1.0 x 10^-2M HCl
HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH
pH = -log[H+]
pH = -log(1.0x10^-2)
pH = 2
b. 1.0 x 10^-3M HNO3
HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.
pH = -log[H+]
pH = -log(1.0 x 10^-3)
pH = 3
c. 1.0 x 10^-1M HI
Like the previous parts, HI is a strong acid
pH = -log[H+]
pH = -log(0.10)
pH = 1
d. HB isn't an element, nor is it a compound so that would be unanswerable.
