The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
coal
Which of these can be mined from earth and used as an energy source?
A.Get rid of the waste and extra fluid
B.Ureters
C.Coal
D.Bladder
ans is coal
Activity series of metals: K,Na,Mg,Al,Zn,Fe,Cu,Ag. Metals on the left are more reactive than metals on the right. For example Zn is more reactive than Fe and can displace him.
Reaction than can occur is: <span>CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s).</span>
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:
Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g
The percent yield of chloro-ethane in the reaction is 82.98%.
Answer:
C
Explanation:
The oxidation number of Sulphur in SO4^2- is;
x + 4(-2) = -2
x - 8 = -2
x = -2 + 8
x = 6
Now,
the oxidation number of sulphur in H2SO3 is
2 (1) + x + 3(-2) = 0
2 + x -6 = 0
-4 + x = 0
x = 4
Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.
