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a chemist adds 240.0 ml of a $7.04\times10^{-5}$ mm zinc oxalate ($\tex…

Question

a chemist adds 240.0 ml of a $7.04\times10^{-5}$ mm zinc oxalate ($\text{znc}_{2}\text{o}_{4}$) solution to a reaction flask. calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. round your answer to 3 significant digits.

Explanation:

Step1: Convert volume and concentration units

First, convert the volume from \(mL\) to \(L\): \(V = 240.0\space mL=240.0\times10^{- 3}\space L = 0.2400\space L\).
Convert the concentration from \(mM\) to \(M\): \(C = 7.04\times10^{-5}\space mM = 7.04\times10^{-8}\space M\) (since \(1\space mM=10^{-3}\space M\)).

Step2: Calculate the number of moles

Use the formula \(n = C\times V\).
\(n=(7.04\times 10^{-8}\space mol/L)\times(0.2400\space L)=1.6896\times10^{-8}\space mol\).

Step3: Calculate the molar mass of \(ZnC_{2}O_{4}\)

The molar mass of \(Zn\) is \(M_{Zn}=65.38\space g/mol\), \(C\) is \(M_{C} = 12.01\space g/mol\), and \(O\) is \(M_{O}=16.00\space g/mol\).
\(M = 65.38+(2\times12.01)+(4\times16.00)=153.4\space g/mol\).

Step4: Calculate the mass in grams

Use the formula \(m=n\times M\).
\(m=(1.6896\times 10^{-8}\space mol)\times(153.4\space g/mol)=2.592\times10^{-6}\space g\).

Step5: Convert grams to micrograms

Since \(1\space g = 10^{6}\space\mu g\), \(m=(2.592\times10^{-6}\space g)\times(10^{6}\space\mu g/g)=2.59\space\mu g\).

Answer:

\(2.59\space\mu g\)