A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 28 s for 1.0 L of O2 gas to effuse.

Part A

Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.) Express your answer using two significant figures.

648g/mol

Explanation:

Let the rate of effusion of the unknown gas be R1 and that of O2 be R2

For the unknown gas:

Volume = 1L

Time = 125sec

R1 = volume / time

R1 = 1/125

R1 = 0.008L/s

For O2:

Volume = 1L

Time = 28sec

R2 = volume / time

R2 = 1/28

R2 = 0.036L/s

Now, we can easily find the molar mass of the unknown gas using the Graham's law equation as shown below:

R1/R2 = √(M2/M1)

R1 = 0.008L/s

R2 = 0.036L/s

M1 (molar Mass of unknown gas) =?

M2 (molar Mass of O2) = 16x2 = 32g/mol

R1/R2 = √(M2/M1)

0.008/0.036 = √(32/M1)

Take the square of both sides

(0.008/0.036)^2 = 32/M1

Cross multiply to express in linear form

(0.008/0.036)^2 x M1 = 32

Divide both side by (0.008/0.036)^2

M1 = 32/ (0.008/0.036)^2

M1 = 648g/mol

Therefore, the molar mass of the gas is 648g/mol

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.