1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
Answer:
PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.
PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.
PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.
Answer:
Explanation:
The half-cell reduction potentials are
Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E° = -0.447 V
To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.
The anode is the electrode at which oxidation occurs.
The equation for the oxidation half-reaction is
Answer:
B. the atomic weight of the element
Explanation:
since it talks about mass i think weight
Answer:
I pretty sure it's the pH level 7
