Answer:
the catalyst is the two gray dots
Explanation:
The reaction equation will be as follows.

Hence, the expression for
is as follows.
![K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH_%7B2%7DSO%5E%7B-%7D_%7B4%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BH_%7B3%7DAsO_%7B4%7D%5D%7D)
Let us assume that the concentration of both
and
is x.

x = 0.01118034
This means that the concentration of
is 0.01118034.
Since, we know that the relation between pH and concentration of hydrogen ions is as follows.
pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
= 
= 1.958
Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.
Explanation:
The given data is as follows.
Air is at
and 14.6 psia.
= 0.00015 ft, Flow rate, (Q) = 48000 
(a) Formula to calculate hydraulic radius
is as follows.

= 
=
ft
Formula for equivalent diameter is as follows.

=
=
ft
(b) Formula for velocity floe is as follows.
Q = VA
V = 
=
ft/min
= 24000 ft/min
(c) Formula to calculate Reynold's number is as follows.
= 
=
(as
and
= 0.0443 lb/ft. hr)
= 53742.66 hr/min
As 1 hr = 10 min. So, 
= 3224559.6
(d) Formula to calculate pressure drop
is as follows.

Putting the given values into the above formula as follows.

= 
= 6.238 
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g
C H O
mass 63.13 g 8.830 g 28.03 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.13/12 8.830/1 28.03/16
5.26 8.830 1.75
divide by the smallest number of moles
5.26/1.75 8.830/1.75 1.75/1.75
= 3.01 = 5.04 =1
rounded off to the nearest whole numbers
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O
Q2)
we have to next determine the molecular formula of the compound
molecular formula gives the actual composition of elements in the compound.
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
= 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
= 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄