The first one is some reaction with water even I am studying the same
Hello!
The molarity of the HBr solution is 0,172 M.
Why?
The neutralization reaction between LiOH and HBr is the following:
HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)
To solve this exercise, we are going to apply the common titration equation:
Have a nice day!
An electrolyte<span> is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. </span>
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
<em />
Answer : The correct option is, (B) 
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of unknown gas = 
= rate of effusion of oxygen gas = 
= molar mass of unknown gas = ?
= molar mass of oxygen gas = 32 g/mole
Now put all the given values in the above formula 1, we get:
The unknown gas could be carbon dioxide 
that has approximately 44 g/mole of molar mass.
Thus, the unknown gas could be carbon dioxide
