3. Ampere is the base unit for electric current.
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:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
To be able to write correctly the equilibrium expression of a reaction, we need to know the balanced reaction and the phases of the substances in the reaction. When substances are solid, pure liquid they are not included in the expression. We do as follows:
<span>4KO2(s) + 2H2O(g) = 4KOH(s) + 3O2(g)
K = [O2]^3 / [H2O]^2</span>