Gas y effuses half as fast as o2. what is the molar mass of gas y?

To solve this question, we will use Graham's law which states that:
(R1 / R2) ^ 2 = M2 / M1 where
R1 and R2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.
From the periodic table, we can calculate the molar mass of O2 as follows:
molar mass of O2 = 2*16 = 32 grams

Therefore we have:
R1 / R2 = Ry / RO2 = 1/2
M1 is My we want to get
M2 is molar mass of O2 = 32 grams
Substitute in the above equation to get the molar mass of y as follows:
(1/2) ^2 = (32/My)
1/4 = 32/My
My = 32*4 = 128

Therefore, molar mass of gas y = 128 grams

If the water vapor content stays the same and the temperature drops, then the relative humidity increases. its vise versa.if the water vapor content stats the same and the temperature rises, the relative humidity decreases. so, yes, i'd say that temperature and humidity have a strong relationship.

The net force is 100