Medgar Evers College QUIZ 3 Prof. J. Etienne instruction: Each question is worth
ID: 517713 • Letter: M
Question
Medgar Evers College QUIZ 3 Prof. J. Etienne instruction: Each question is worth 20 points 1. Based on your knowledge of gas laws, how do you now when two variables are inversely proportional to each other? Give an example of two variables that are inverse to each other and two variables that vary directly in proportion to each other. Please give specific examples for such variables, 2. The reading on a Kelvin scale ls 323K. on the Celsius scale the reading would be what? 3. A student did an experiment to calculate percent composition of a mixture of Potassium chlorate and Potassium chloride. At the end of the experiment, student determined that the mixture had 71 percent of Potassium chlorate and 39 percent of potassium chloride. Do you think this data make sense? Explain your reasoning 4. A chemist heated an unknown hydrate, Meso, x Hho, in a crucible. The water was driven off, leaving the anhydrous salt in the crucible. After cooling, the mass of the anhydroussalt and water lost was calculated. The mass of the hydrated salt was 2.5 grams gand the mass of the water lost was 1-1 gram. Is it possible to find the mass of the anhydrous salt? Please show your calculations or explain 5. are the three main variables that are involved in Boyle's law. Name them with their correct variables you must convert to unit when solving a problem on gas law.Explanation / Answer
1) Consider ideal gas law
PV = nRT where R is constant
P is proportional to T
P is proportional to n
where as P is inversely proportinal to V
we know that these are inversely proportional if we keep all other constant as
PV = k where k is constant
P = k/V
now if increase V , p will decrease and if you decrease V, P will decrease
T is inversely proportional to n
2) Tempertaure = 323 K
Kelvin to celcius we have to subtract 273 K
Celcius temperature = 323 - 273 = 50 C
3) No it doesn't make sense because sum of percent cant be more than 100
composition cannot be greater than 100
here given as 71 percent and 39 percent
sum = 71 + 39 = 110 which is not possible
4) Given
mass of hydrated salt = 2.5 g
mass of water lost = 1.1 g
Mass of anhydrous salt = 2.5 g - 1.1 g = 1.4 g Answer
5) three variables involved are P, V ,k where k is constant k = nRT
boyle's law PV = k
P unit is Pa
V unit is m3
k unit is J