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
Answer:
2.5, - 2.5
Explanation:
Given that :
Coordinate of A = 0
and AR = 5 ; distance between A and R
Possible. Coordinate of the midpoint of AR overbar.
Hence for a distance of 5 points between A and R, then R may be located 5 points either side of A, that is to the left of right.
Kindly check attached picture for detailed workings
Between 1.00 & 1.5 grams would be the normal amount
They discovered this since S waves don’t travel through the earth but P waves do. P waves tend to travel faster than S waves in speed, 1&14km/s vs 1&8km/s