Using the titration formula, we get that:
(concentration of HBr)(25)=(0.5)(15.3)
This means the molar concentration of HBr is (0.5)(15.3)/25 = 0.31 M
Answer:
2.8087*10^-12 kJ per mole of reaction (2.8087*10^-12 kJ/mol).
Explanation:
To calculate the energy produced, we need to write a balanced equation for the reaction and determine the change in the masses of the reactants and products. Afterward, we can use the energy equation to determine the energy produced. The balanced equation for the nuclear reaction is shown below:
³₁H + ²₁H ⇒⁴₂He + ¹₀n
The masses of atoms are ³₁H is 3.01605 amu, ²₁H is 2.0140 amu, ⁴₂He is 4.00260 amu, and ¹₀n is 1.008665 amu.
change in mass Δm = (3.01605+2.0140) - (4.00260+1.008665) = 0.0188 amu
Energy produced, E = m*C^2
C is the speed of light = 3*10^8 m/s and 1 amu = 1.66*10^-27 kg
Therefore:
E = 0.0188*1.66*10^-27 * (3*10^8)^2 = 2.8087*10^-12 kJ per mole of reaction.
Therefore, in scientific notation, the energy released is 2.8087*10^-12 kJ/mol
Metals are most likely because they conduct heat well
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
