Answer:
the answer is b Li + Cl2 .....
C—stems and leaves growing upward?
Answer:
element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.
Explanation:
- Group IIA have 2+ valency and two electrons in its valance shell.
- Its Electropositivity is high and have the tendency to donate it two electrons.
- Element of IIA form ionic with most electronegative element.
Examples:
Cu²⁺, Mg²⁺, Sr²⁺ are examples having 2+ valance electron
one of the following is examples of element that have 2+ valence electrons
MgCl₂
Atomic number of Magnesium (Mg) is 12
Electronic Configuration of Mg:
1s², 2s², 2p⁶, 3s²
or
K =2
L = 8
M = 2
So, it have to give its 2 electrons to form a stable compound.
Similarly
Chlorine atomic number is 17
Electronic Configuration of Chlorine:
1s², 2s², 2p⁶, 3s², 3p⁵
or
K =2
L = 8
M = 7
So, it have to gain one electrons to form a stable compound and complete its octet.
So,
Two chlorine atom as a molecule gain 2 electrons from Mg²⁺ atom
So one Mg²⁺ and 2 Cl⁻ atoms form an ionic bond
where in this ionic bond Mg²⁺ transfer its 2 valence electron completely and chlorine molecule accept 2 electrons.
Cl-----Mg------Cl
So the Answer is
element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.
Answer:
Part A:
First, convert molarity to moles by multiplying by the volume:
0.293 M AgNO3 = (0.293 moles AgNO3)/1 L x 1.19 L = 0.349 moles AgNO3
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
