A new molecule called Annaxyl was just discovered with a molecular mass of 25 g/mol and a density of 0.345 g/mL. If a super-scientist named Anna decided to add 20.0 mL of Annaxyl to 3.00 L of water, what is the concentration (mol/L) of Annaxyl?
A new molecule called Annaxyl was just discovered with a molecular mass of 25 g/mol and...
An unknown amount of a compound with a molecular mass of 274.11 g/mol 274.11 g/mol is dissolved in a 10 mL volumetric flask. A 1.00 mL aliquot of this solution is transferred to a 25 mL volumetric flask, and enough water is added to dilute to the mark. The absorbance of this diluted solution at 325 nm 325 nm is 0.523 0.523 in a 1.000 cm cuvet. The molar absorptivity for this compound at 325 nm 325 nm is ?...
An unknown amount of a compound with a molecular mass of 294.46 g/mol is dissolved in a 10 mL volumetric flask. A 1.00 mL aliquot of this solution is transferred to a 25 mL volumetric flask, and enough water is added to dilute to the mark. The absorbance of this diluted solution at 343 nm is 0.456 in a 1.000 cm cuvette. The molar absorptivity for this compound at 343 nm is €343 = 5721 M-?cm-1. What is the concentration...
An unknown amount of a compound with a molecular mass of 284.04 g/mol is dissolved in a 10 mL volumetric flask. A 1.00 mL aliquot of this solution is transferred to a 25 mL volumetric flask, and enough water is added to dilute to the mark. The absorbance of this diluted solution at 347 nm is 0.472 in a 1.000 cm cuvet. The molar absorptivity for this compound at 347 nm is ?347=6023 M−1cm−1. What is the concentration of the...
An unknown amount of a compound with a molecular mass of 270.93 g/mol is dissolved in a 10-mL volumetric flask. A 1.00-mL aliquot of this solution is transferred to a 25-mL volumetric flask and e water is added to dilute to the mark. The absorbance of this diluted solution at 339 nm is 0.517 in a 1 cm cuvette. The molar absorptivity for this compound at 339 nm is esso- 5835 M-1cm-1 (a) What is the concentration of the compound...
An unknown amount of a compound with a molecular mass of 259.45 g/mol is dissolved in a 10-mL volumetric flask. A 1,00-mL aliquot of this solution is transferred to a 2 L volumetric flask and o the mark. The absorbance of this diluted solution at 347 nm is 0.468 in a 1 cm cuvette. The molar (a) What is the concentration of the compound in the cuvette? Number b) What is the concentration of the compound in the 10-ml flask?...
How much urea (CH4N2O; molecular weight: 60.056 g/mol) in grams do I need to add to 1L water to have a final concentration of 20 mg/L Total Nitrogen as Nitrogen or TN-N. Nitrogen molecular weight is 14.0067 g/mol.
A newly synthesized drug (molecular mass = 322 g/mol) was tested for its optical activity. To do this, 15.0 grams of the drug was dissolved in water and brought to a volume of 12 mL. This solution was placed in a 7.50 cm cell and the optical rotation was found to be -31.0°. What is the specific rotation for this new drug?
A 3.80-g sample of an unknown compound containing only C, H, and O combusts in an oxygen rich environment. When the products have cooled to 20.0 degree C at 1 bar, there are 4.96 L of CO_2 and 2.45 mL of H_2O. The density of water at 20.0 degree C is 0.998 g/mL. What is the empirical formula of the unknown compound? If the molar mass is 168.2 g/mol, what is the molecular formula of the compound?
1) An aqueous solution containing 15.9 g of an unknown molecular (nonelectrolyte) compound in 103.0 g of water was found to have a freezing point of -1.5 ∘C. Calculate the molar mass of the unknown compound. b) Determine the required concentration (in percent by mass) for an aqueous ethylene glycol (C2H6O2) solution to have a boiling point of 109.3 ∘C. c) What mass of salt (NaCl) should you add to 1.90 L of water in an ice cream maker to...
Concentrated hydrochloric acid (HCl) (molecular mass 36.5 g/mol) has a molarity of 3.6M. What volume of concentrated HCl would you need to add to prepare a 20% (w/v) solution of HCl with a final volume of 500 mL, I have calcualted that 3.6mol/L x 36.5g/mol =131.40g/L. Then to get the 100ml value and percentage: 131.40/1000 x 100 =13.14g/100ml therefore 13.14%. 20% of this would be 2.628g. Is this the answer?