Calicheamicin gamma-1, C55H74IN3O21S4, is one of the most potent antibiotics known: One molecule kills one bacterial cell. Describe how you would (carefully!) prepare 25.00 mL of an aqueous calicheamicin gamma-1 solution that could kill 1.0×108 bacteria, starting from a 5.00×10−9M stock solution of the antibiotic. Complete the explanation. The 25.00 mL of antibiotic solution needs to contain a minimum of 1.0×108 The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M . molecules of the drug. Calculate the moles The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M . of drug this represents. The concentration of the stock solution is 5.00×10−9 The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M . . Then, L stock solution = mol drug/M solution; L = mol/5.00×10−9 The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M . M. Calculate the volume of the antibiotic solution needed to kill the bacteria from Part A.
Calicheamicin gamma-1, C55H74IN3O21S4, is one of the most potent antibiotics known: One molecule kills one bacterial cell.
Describe how you would (carefully!) prepare 25.00 mL of an aqueous calicheamicin gamma-1 solution that could kill 1.0×108 bacteria, starting from a 5.00×10−9M stock solution of the antibiotic. Complete the explanation.
The 25.00 mL of antibiotic solution needs to contain a minimum of 1.0×108
The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M .
molecules of the drug. Calculate the moles
The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M .
of drug this represents. The concentration of the stock solution is 5.00×10−9
The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M .
. Then, L stock solution = mol drug/M solution; L = mol/5.00×10−9
The 25.00 m L of antibiotic solution needs to contain a minimum of 1.0 × 10 8 molecules of the drug. Calculate the moles of drug this represents. The concentration of the stock solution is 5.00 × 10 − 9 . Then, L stock solution = m o l drug / M solution; L = m o l / 5.00 × 10 − 9 M .
M.
Calculate the volume of the antibiotic solution needed to kill the bacteria from Part A.
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