Election current because voltage is a measurement, information doesn't apply to all electrical devices and the wires within are usually copper bc it conducts and hardly ever will the wires be anything different because copper is cheap
Answer : The correct options are, 
and 
Explanation :
Single displacement reaction : It is a type of chemical reaction in which the more reactive element displaces the less reactive element.
Option A reaction : 
It is an example of double displacement reaction because in this reaction a positive cation and a negative anion of the two reactants exchange their places to form two new products.
Option B reaction :
It is an example of single displacement reaction.
Option C reaction : 
It is an example of combination reaction because in this reaction two reactants react to give a single product.
Option D reaction : 
It is an example of decomposition reaction because in this reaction a single reactant decomposes into two or more products.
Option E reaction :
It is an example of single displacement reaction because in this reaction the most react element, aluminium displaces the less reactive element, hydrogen.
Hence, the options B and E are single displacement reactions.
Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10
