The molar extinction coefficient is 15,200
.
The formula to be used to calculate molar extinction coefficient is -
A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.
The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.
Rewriting the formula as per molar extinction coefficient -
ξ = 
ξ = 
Performing multiplication in denominator to find the value of molar extinction coefficient
ξ =
Performing division to find the value of molar extinction coefficient
ξ = 15,200 
Hence, the molar extinction coefficient is 15,200
.
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Answer:
1.22 L of carbon dioxide gas
Explanation:
The reaction that takes place is:
- CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
First we <u>determine which reactant is limiting</u>:
- Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
- Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl
So HCl is the limiting reactant.
Now we calculate the moles of CO₂ produced:
- 0.05 mol HCl *
= 0.05 mol CO₂
Finally we use PV=nRT to <u>calculate the volume</u>:
- T = 25 °C ⇒ 25 + 273.16 = 298.16 K
1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Answer:
an atom cannot be broken down
Answer:
This description needs a unit describing the system of measurement.