Answer:
The right statement is "The percent transmittance of light, %T, will be higher for Sample 1 compared to Sample 2."
Explanation:
The Beer-Lambert Law states that the absorbance is directly proportional to the concentration of the substance.
Absorbance = εLc
ε: molar absorptivity coeficient
L: optical path length
c: molar concentration
Therefore, if ε and L are the same, the higher the concentration, the higher the absorbance and the lower the transmitance.
Since sample 1 has a higher concentration of CuSO₄ than sample 2, sample 1 has a higher absorbance and lower percent transmittance.
Reaction is NH4OH <-> NH4+ OH- (note this is reversible)
Draw up an ICE table
Let x be equilibrium conc of OH- assume init conc of OH is 0M and init conc of NH4+ is 0M also. Init conc of NH4OH is 0.1M so equilibrium conc will be 0.1-x.
%dissociation = x/0.1-x * 100%
1 = 100x/0.1-x
0.1-x = 100x
101x = 0.1
x = 0.0009901
pOH = -log(0.0009901) = 3.00
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
Answer:
D
Explanation:
It is D because once youve put together the ingredients and baked it you can not go in and taake out every ingredient and put it back to how it was before (Please also try to help me with my question it it about scaled copies, Please and thank you!)
