Raising of the boiling point is a colligative property. That means that it depends on the number of particles dissolved. The greater the number of particles the greater the increase in the boiling point. So, you can compare the effect of these solutes in the increase of the boiling point by writing the chemical equations and comparing the number of particles dissolved: 1)ionic lithium chloride, LiCl(s) --> Li(+) + Cl (-) => 2 ions; 2) ionic sodium chloride, NaCl(s) --> Na(+) + Cl(-) => 2 ions; 3) molecular sucrose, C12H22O11 (s) ---> C12H22O11(aq) => 1 molecule; 4) ionic phosphate, Na3PO4 --> 3Na(+) + PO4 (3-) => 4 ions; 5) ionic magnesium bromide, MgBr2 --> Mg(2+) + 2 Br(-) => 3 ions. <span>So, ionic phosphate produces the greatest number of particles and it will cause the greatest increase of the boiling point.</span><span />
<u>Given:</u>
Moles of He = 15
Moles of N2 = 5
Pressure (P) = 1.01 atm
Temperature (T) = 300 K
<u>To determine:</u>
The volume (V) of the balloon
<u>Explanation:</u>
From the ideal gas law:
PV = nRT
where P = pressure of the gas
V = volume
n = number of moles of the gas
T = temperature
R = gas constant = 0.0821 L-atm/mol-K
In this case we have:-
n(total) = 15 + 5 = 20 moles
P = 1.01 atm and T = 300K
V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L
Ans: Volume of the balloon is around 488 L
The % yield of Ca(OH)₂ : 62.98%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
Reaction
CaO + H₂O ⇒ Ca(OH)₂
mass CaO= 4.2 g
mol CaO(MW=56,0774 g/mol) :
mol Ca(OH)₂ based on mol CaO
mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075
mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical
% yield :
For example...
You have solution with [H+] = 0,01M
>>>> pH = -log[H+] = -log0,01 = 2
And you increase the [H+] by 10x ---> 0,01×10 = 0,1M
>>>> pH = -log0,1 = 1
○ pH decrease by 2x
○ pH is more acidic
