Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
Answer:
ethyl acetate layer
Explanation:
This question tests your knowledge of the principle of solvent extraction.
In solvent extraction, there is always an organic layer and an aqueous layer. The ethyl acetate is the organic layer while the sodium bicarbonate is the aqueous layer.
A brominated aromatic compound will be extracted into the organic layer (ethyl acetate layer).
Mghcl14 + H2 = mghclh 16 work needed to solve this problem
Answer:
The concentration c is equal to Ka
Explanation:
The acid will ionize as observed in the following reaction:
HA = H+ + A-
H+ is the proton of the acid and A- is the conjugate base
. The equation to calculate the Ka is as follows:
Ka = ([H+]*[A -])/[HA]
Initially we have to:
[H+] = 0
[A-] = 0
[HA] = c
During the change we have:
[H+] = +x
[A-] = +x
[HA] = -x
During balance we have:
[H+] = 0 + x
[A-] = 0 + x
[HA] = c - x
Substituting the Ka equation we have:
Ka = ([H+]*[A-])/[HA]
Ka = (x * x)/(c-x)
x^2 + Kax - (c * Ka) = 0
We must find c, having as [H+] = 1/2c. Replacing we have:
(1/2c)^2 + (Ka * 1/2 * c) - (c * Ka) = 0
(c^2)/2 + Ka(c / 2 - c) = 0
(c^2)/2 + (-Ka * c/2) = 0
c^2 -(c*Ka) = 0
c-Ka = 0
Ka = c
