For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as
A’ = 0.16138
<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>
Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069
Generally, the equation for the Beer’s law is mathematically given as
A = ε*c*l
0.8069 = ε*c*(5.00 )
ε*c = 0.16138 cm-1
then for when ε*c is constant
l’ = 1.00
A’ = (0.16138 cm-1)*(1.00 cm)
A’ = 0.16138
In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is
A’ = 0.16138
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Answer:
The minimum pressure should be 901.79 kPa
Explanation:
<u>Step 1: </u>Data given
Temperature = 25°C
Molarity of sodium chloride = 0.163 M
Molarity of magnesium sulfate = 0.019 M
<u>Step 2:</u> Calculate osmotic pressure
The formula for the osmotic pressure =
Π=MRT.
⇒ with M = the total molarity of all of the particles in the solution.
⇒ R = gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 25 °C = 298 K
NaCl→ Na+ + Cl-
MgSO4 → Mg^2+ + SO4^2-
M = 2(0.163) + 2(0.019 M)
M = 0.364 M
Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)
Π = 8.90 atm
(8.90 atm)(101.325 kPa/atm) = 901.79 kPa
The minimum pressure should be 901.79 kPa
Radon is a colorless, odorless, radioactive gas. It is also a leading cause of lung cancer. Ventilation is essential.
For every, 3 Br- ions, 1 Al3+ ion reacts to form AlBr3.
Convert 16.2g of aluminum to moles:
16.2g Al / 27.0g per mol = 0.60 mols.
Based on the above ratio, 0.60 mols of Al will react with 1.8 mols of Br.
Convert 1.8 mols of Br to its mass:
1.8 mols Br × 79.9g per mol = 143.82g of Br.
Explanation:
It is given that the total volume is (10 mL + 60 mL) = 70 mL.
Also, it is known that 
= 
Where, 
= total volume
= initial volume
Therefore, new concentration of 
= 
= 
= 0.43 M
New concentration of NaOH =
= 
= 0.14 M
So, the given reaction will be as follows.
Initial: 0.43 0.14 0
Change: -0.14 -0.14 0.14
Equilibrium: 0.29 0 0.14
As it is known that value of 
= 4.74
Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.
pH = ![pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 
= 4.74 + (-0.316)
= 4.42
Therefore, we can conclude that the pH of given reaction is 4.42.
