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:
i believe that the answer is c
Explanation:
I think its c because because its used as a bed rock layer
Answer:
See below
Explanation:Plot the known concentrations and adsorbance data. Draw a best fit line through thwe points. When the absorbance of a solution of unknown concentration (but same substance) is determined, find the concentration from the line at that absorption value. See attached graph.
E.g., an sample of the same substance had an absorbance of 0.35. Find that on the x scale and then determine the concentration that would be required to produce that level of absorbance. 0.483M in this case.
Answer:
Potassium dihydrogen phosphate is a potassium salt in which dihydrogen phosphate(1-) is the counterion. ... It is a source of phosphorus and potassium as well as a buffering agent. It can be used in fertilizer mixtures to reduce escape of ammonia by keeping pH low.
Explanation:
