Find the moles of CaO
divide mass (2.0g) by the RFM which is 56 (Ca is 40 add that to O which is 16 making 56) this gives 0.0356 moles.
Find the theoretical mass by multiplying the moles of CaO (which is 0.0356 as there are no balancing number making the ratio 1:1) by the RFM of Ca(OH)2 which is 74 (40+16+16+1+1)
74 (Ca(OH)2 RFM) x 0.0357 (CaO moles) = 2.6g which is the theoretical mass of Ca(OH)2
Find percentage yield by dividing the actual mass of Ca(OH)2 by the theoretical and then x100 this Should give you 82.3%
Molality is a measure of concentration that relates the moles of solute to the kilograms of solvent, it is described by the following equation:

We are given the molality(2.50m) and kilograms of solvent(0.500kg), so we solve for moles of solute from the equation:


To make a 2.50molal NaOH solution would be needed 1.25moles of solute
Answer: 1.25 moles
Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :

act as an oxidizing agent.
The oxidation number of N in
is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in
is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means
itself get reduced to
and it can act as an oxidizing agent.
act as a reducing agent.

The oxidation number of O in
is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in
is Zero (o).
Now, we conclude that the oxidation number increases this means
itself get oxidized to
and it can act as reducing agent.
Explanation:
30 lb is 480 ounces
34 mi/second is 54.718 kilometre/ second
455 lb/ gal is 54521.024 grams / litre
50 cl is 500 millilitres
55nm is 5.5 × 10^-6 centimetre