For the neutralization process: an acid acts as a donor and donates protons to the base. On the other hand, the base acts as an acceptor and accepts the transferred protons. In a nutshell, neutralization is mainly proton transfer process.
As for the redox process: the oxidized material usually transfers electrons to the reduced material. In a nutshell, redox is mainly electron transfer process.
The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...
Explanation:
By way of example, chemists and physicists habitually use
P
saturated vapour pressure
...where
P
SVP
is the vapour pressure exerted by liquid water. At
100
∘
C
,
P
SVP
=
1
⋅
a
t
m
. Why?
Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here
1
⋅
a
t
m
) and temperature, here
100
∘
C
, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...
At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of
saturated vapour pressure
are available.
Answer:
13.94moles of Na₂O
Explanation:
The balanced reaction expression is given as:
4Na + O₂ → 2Na₂O
Given parameters:
Number of moles of O₂ = 6.97moles
Unknown:
Number of moles of Na₂O
Solution:
To solve this problem;
1 mole of O₂ will produce 2 moles of Na₂O ;
6.97 moles of O₂ will produce 6.97 x 2 = 13.94moles of Na₂O
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
