<span>I believe the correct 2nd reaction is:</span>
cof2(g)⇌1/2 co2(g)+1/2 cf4(g)
where we can see that it is exactly one-half of the
original
Therefore the new Kp is:
new Kp = (old Kp)^(1/2)
new Kp = (2.2 x 10^6)^(1/2)
<span>new Kp = 1,483.24 </span>
Which property of gas affects the gas by change in that property ?
All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:
</span><span>ΔT = -K (m) (i)
</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.
<span>
K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.
</span><span>
m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.
</span><span>
i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>
Answer:
Li(s) + H20(l) --> LiOH(aq) + H(g)
Explanation:
it is already balanced
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
