Chemistry, 24.10.2019 23:43 letsbestupidcx2314
We will calculate the amount of acid to use in each titration. assume thaqt you are using 0.512 m naoh (aq). a good volume of naoh to use per titration is 15 ml. from this molarity and volume, the moles of naoh can be calculated. since the unknown acid is monoprotic, this also equals the number of moles of acid to use. a typical molar mass for an unknown acid in this experiment is 380 g/mol. using this molar mass, calculate the mass (in grams) of unkmown acid you should use per titration. round to 2 sig figs.
-suppse that 14.0 ml of 0.0512 m naoh were required to titrate a sample of uknown acid. how many moles of naoh were used?
-assuming that the unknown acid sample in question 1 had a mass of 0.177 g what is the molar mass of the unknown acid?
Answers: 1
Chemistry, 22.06.2019 05:30
Match the following vocabulary terms to their definitions. 1. amount of energy required to change 1 gram of material from the solid to the liquid state at its melting point 2. a measure of the kinetic energy of the particles of a substance 3. the amount of heat energy required to raise the temperature of 1 gram of liquid water from 14.5°c to 15.5°c 4. amount of energy required to change 1 gram of material from the liquid to the gaseous state at its boiling point 5. the amount of energy required to change 1 gram of a substance 1°c a. temperature b. latent heat of vaporization c. latent heat of fusion d. calorie e. specific heat
Answers: 1
Chemistry, 22.06.2019 21:50
Liquid from a brewery fermentation contains 10% ethanol and 90% water. part of the fermentation product (50,000 kg/h) is pumped to a distillation column on the factory site. under current operating conditions, a distillate of 45% ethanol and 55% water is produced from the top of the column at a rate of one-tenth that of the feed. what is the composition of the waste "bottoms" from the still?
Answers: 2
We will calculate the amount of acid to use in each titration. assume thaqt you are using 0.512 m na...
Mathematics, 23.05.2020 17:59
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