calcium,phosphorus,potassium,and sulfer
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer:
Explanation:
Coinage metals -
The group 11 elements , i.e. , Copper , Silver , Gold are called the coinage metals .
These metals are quite soft and can be easily molded to form coins . This property is called as the property of malleability .
And due its soft nature , it can easily be used to make jewelry ,
Since , making jewels require the metal to be soft and flexible ,
Hence , gold is suited the best for this .
Answer:
The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.
Explanation:
Step 1:
...[1]
Step 2 :
..[2]
Adding 6 × [1] and [2]:
we get :
Energy released on formation of 5 moles of acrylic acid = 2,626 kJ
Energy released on formation of 1 mole of acrylic acid:
The answer is A because I did this before
