Answer:
The pH is equal to 4.41
Explanation:
Since HClO is a weak acid, its dissociation in aqueous medium is:
HClO ⇄ ClO- + H+
start: 0.05 0 0
change -x +x +x
balance 0.05-x x x
As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.
the acidity constant when equilibrium is reached is equal to:
The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:
clearing the x and calculating its value we have:
the pH can be calculated by:
Mixture. Distilled Water may be a pure substance, however, there are many minerals and such that are in tap water, making it a mixture.
Answer:
0.550
Explanation:
The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.
A = ε . <em>l</em> . c
where,
ε: absorptivity of the species
<em>l</em>: optical path length
A 45 mM phosphate solution (solution A) had an absorbance of 1.012.
A = ε . <em>l</em> . c
1.012 = ε . <em>l</em> . 45 mM
ε . <em>l</em> = 0.022 mM⁻¹
We can find the concentration of the second solution using the dilution rule.
C₁ . V₁ = C₂ . V₂
45mM . 11mL = C₂ . 20.0 mL
C₂ = 25 mM
The absorbance of the second solution is:
A = (ε . <em>l</em> ). c
A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L