Answer:
See explanation
Explanation:
Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.
The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).
The formation of the hydride ion completes the 1s orbital.
The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.
Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).
Nomenclature:
In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.
Name of Formate:
Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.
E.g.
HCOOH (formic acid) → HCOO⁻ (formate) + H⁺
H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺
Formal Charges:
Formal charges are calculated using following formula,
F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]
For Oxygen:
F.C = [6] - [6 + 2/2]
F.C = [6] - [6 + 1]
F.C = 6 - 7
F.C = -1
For Sodium:
F.C = [1] - [0 + 0/2]
F.C = [1] - [0]
F.C = 1 - 0
F.C = +1
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
ΔS =S(products) -S(reactants)
Where ΔS is the change of entropy in a reactions
a. ΔS = (2) - (2+1) = -1
b. ΔS = (1+1) -(1) = 1
c. ΔS = (1+2) - (1) = 2
d. ΔS = (2) - (2+1) = -1
e. ΔS = (1) - (1) = 0
ΔS is negative for reaction a. and d.
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
