Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
Matter is anything that occupies space and has mass.
There are a few ways to do this. The best way I can think of is to start out the paint with no additives and have that as a control. You can test that on whatever your painting or do it in a lab with heat lamps and lights if that is a possible option. However you decide to do the testing, after starting with the control test, add different additives and see which ones fades less the original without any additives. This is the best scenario for a simple yet informative test.
Hope that helps!
Answer:
3.6 gallons are equal to 56 cups
A reaction is first order. If the initial reactant concentration is 0.0200 M, and 25.0 days later the concentration is 6.25 x 10-4 M, then its half-life is: