To know the acidity of a
solution, we calculate the pH value. The formula for pH is given as:
<span>pH = - log [H+] where H+ must be in Molar</span>
We are given that H+ = 3.25 × 10-2 M
Therefore the pH is:
pH = - log [3.25 × 10-2]
pH = 1.488
Since pH is way below 7, therefore the solution
is acidic.
To find for the OH- concentration, we must
remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,
[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]
[OH-] = 10^-14 / 3.25 × 10-2
[OH-] = 3.08 × 10-13 M
Answers:
Acidic
[OH-] = 3.08 <span>× 10-13 M</span>
Answer: 253.8
Explanation:
The molar mass of iodine is 126.904. Multiply that by two and you get approximately 253.8 grams in two moles.
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
Answer:
Lead
that can be found in the periodic table of elements
Explanation:
Answer:
Explanation:
Density is
mass / volume = d
Mass:
840g
Volume:
7 cm x 4 cm x 10 cm = 280 cm^3
840g / 280 cm^3 = 3 g/cm^3
