Waves with higher frequencies have shorter wavelengths, and lower frequencies have longer wavelengths
B) They have protons that are identical to the protons of all other elements
Answer and Explanation:
Aspirin is odorless, but when left exposed to air in the environment, it gradually hydrolyzes into salicylic acid and acetic acid as that is the precursor for synthesizing Aspirin.
Using this hydrolyzed aspirin for titration would not be advised, because it would affect the reading of the titration. Ordinarily, apsirin is a weak acid and direct titration of aspirin is problematic because it hydrolyzes pretty fast to salicylic acid— leading to an unwanted side reaction which may or may not go to completion. Therefore, excess base must be added and heat is supplied to the mixture so that neutralization and hydrolysis are complete. The remaining base is then titrated. This is called back titration.
Now, in back titration, instead of using solution whose concentration is expected to be known, we rather use excess volume of reactant which has been left over after the completion of a reaction with the analyte.
In this case, we use an alkali, preferably NaOH (1.0 mol/dm³). Te unused NaOH remaining after the hydrolysis is titrated against a standard HCl (0.1 mol/dm³). Then from the reaction equation of the aspirin and sodium hydroxide, the amount of NaOH required for the hydrolysis can be calculated.
Answering whether the titration goes up or down, it would be observed that the titration reading would GO DOWN because the exposed aspirin used has experienced some form of hydrolysis before it was used for titration, so the hydrolysis reaction it would undergo with acetyl-salicylic acid would be minimal, and this would affect the titration reading.
But if the aspirin wasn't left exposed to the environment, the reading would go up since more hydrolysis would take place in this case.
Loschmidt's constant is referred to as Avogadro's number (they are the same). Their values are 6.02 times 10 to the -34 power.
Answer:
42.6 % CHCl₃, 13.83 % C₃H₆O and 43.5 % CH₃COBr
Explanation:
This is a solution with 3 compounds:
96.2 g of CHCl₃
31.2 g of C₃H₆O
98.1 g of CH₃COBr
Total mass of solution is : 96.2g + 31.2 g + 98.1g = 225.5 g
Percent by mass is, the mass of the compounds in 100 g of solution
Let's prepare the rule of three
225.5 g of solution have ____ 96.2 g ___ 31.2 g _____98.1 g
100 g of solution would have ____
(100 . 96.2) / 225.5 = 42.6
(100 . 31.2) / 225.5 = 13.83
(100 . 98.1) / 225.5 = 43.5
