Answer:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it.
Crude appearance.
18 18 0
F --> O + e
9 8 1
Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts.
The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons.
Followed by a right arrow to indicate the direction the reaction is going.
Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons.
Followed by a plus sign to indicate more.
Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
Answer:
for what school? It's different for all sadly :(
Explanation:
Answer: HBr has the lowest rate of effusion at a given temperature.
Explanation: The effusion rate usually increases with increase in temperature because the kinetic energy of the gaseous molecules increases. But it was not true for gases having heavier mass. This was explained by Graham's Law.
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight.

We are given different gases with different Molecular masses. The gas having larger Molecular mass will have the lowest rate of effusion.
Mol. Mass of
= 80 g/mol
Mol. Mass of
= 16 g/mol
Mol. Mass of
= 17 g/mol
Mol. Mass of HBr = 81 g/mol
Mol. Mass of HCl = 36 g/mol
As, Mol. mass of HBr is the highest, so its rate of effusion will be the lowest.