Answer: i beleive it is fixation in edge 2020
Explanation:
The wt% of KOH = 45%
This implies that there is 45 g of KOH in 100 g of the solution
Density of the solution is given as 1.45 g/ml
Therefore, the volume corresponding to 100 g of the solution is
= 100 g * 1 ml /1.45 g = 68.97 ml = 0.069 L
Now concentration of the concentrated KOH solution is:
Molarity = moles of KOH/vol of solution
= (45 g/56.105 g.mol-1)/0.069 L = 11.6 M
Thus,
Initial KOH concentration M1 = 11.6 M
Initial volume = V1
Final concentration M2 = 1.20 M
Final volume V2 = 250 ml
M1*V1= M2*V2
V1 = M2*V2/M1 = 1.20*250/11.6 = 25.9 ml = 26 ml
Answer:
the Molar heat of Combustion of diphenylacetylene =
Explanation:
Given that:
mass of diphenylacetylene = 0.5297 g
Molar Mass of diphenylacetylene = 178.21 g/mol
Then number of moles of diphenylacetylene =
=
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene = Heat absorbed by + Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene = msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene = 20598.87 J
Heat liberated by 1 mole of diphenylacetylene will be =
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene =
Answer:
yahh
Explanation:
a precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed.