Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
-573.67
Explanation:
whenever energy is released in a chemical reaction, we would then expect the delta H of the reaction to be negative because the reaction is an exothermic reaction.
now we have that 2.81 moles of fuel when it combusts would releases 1612kJ of energy
thus, 1 mole will release 1612/2.81 = -573.67kJ of heat
Therefore the delta H of the reaction = -573.67 kJ/mol
Answer: The volume of 
required is 25.0 ml
Explanation:
According to the neutralization law,
where,
= basicity 
= 1
= molarity of solution = 2.00 M
= volume of solution = 50.0 ml
= acidity of = 1
= molarity of solution = 4.00 M
= volume of solution = ?
Putting in the values we get:
Therefore, volume of required is 25.0 ml
Concentration of the reactant,pressure,surface
area of the reactant and temperatur
