Calculate the extinction coefficient where the concentration is in mg/ml and the path length is 1 cm. What dilutions of the stoc
k are each of the prepared solutions (i.e., 1/x)?
The molecular weight of A is 290 g/mole.
Re-calculate the extinction coefficient with the concentration in mM. Note that the newly calculated extinction coefficient will contain an mM-1 term.
The molecular weight of A is 290 g/mole.
Re-calculate the extinction coefficient with the concentration in mM. Note that the newly calculated extinction coefficient will contain an mM-1 term.
1 answer:
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Answer:
76.0%
Explanation:
Let's consider the following reaction.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
At equilibrium, the equilibrium constant Kp is:
Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm
We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.
From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:
The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:
Answer:
768g
Explanation:
We can use to formula . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and
is the half life. Plugging in, we get the answer above.