Answer:
70.77 g/mol is the molar mass of the unknown gas.
Explanation:
Effusion is defined as rate of change of volume with respect to time.
Rate of Effusion=
Effusion rate of oxygen gas after time t = 
Molar mass of oxygen gas = M = 32 g/mol
Effusion rate of unknown gas after time t = 
Molar mass of unknown gas = M'
The rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:



M' = 70.77 g/mol
70.77 g/mol is the molar mass of the unknown gas.
Answer:
3.01 ·10↑22
Explanation:
First you want to convert the grams of Glucose to moles of Glucose.

Next find the formula units of glucose.
.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =
5.01 ·10↑21 Formula Units of Glucose
Now multiply the formula units of glucose by the amount of each element in the molecule.
So for Carbon:
6carbon · 5.01 · 10↑21 = 3.01 · 10↑22
(please mark as Brainliest)
called the Avogadro number
N(A)= 6.02 x 10^23 mol^-1
1 mole of SO3 will contain 6.02 x 10^23 mol^-1 of SO3 molecules.
thus, 1.14moles will contain;
= 1.14mol x [3mol O/1mol SO3] x [6.02 x 10^23
atoms O/1mol O]
= 2.05884 x 10^24 oxygen atoms
= 1.14mol x [1mol S/1mol SO3] x [6.02 x 10^23
atoms O/1mol O]
= 6.8628 x 10^23 sulfur atoms
hope this helps:-)
Answer:
Option B: It remains unused during the reaction
Explanation:
Just took the test and got it right.