Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature
Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules
The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).
In studying chemistry, one should learn how to read the periodic table and periodic trends in the table.. Of course, you should also know the abbreviations of each element in it so that you can identify what element is in chemical structures. Memorizing and knowing it by heart is the best way. You should also study what we call chemical nomenclature which is naming combinations of elements. Elements when combined may form ionic compounds, bonds or acids. Knowing what element reacts with another will also be necessary that's why the concept of chemical reactions must be further studied.
Answer:
The correct answer is - 29.45 / 100 x 25.6 = 7.5392 grams
Explanation:
It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:
mass of Ca = total amount of sample*percentage of calcium in sample /100
M of Ca =25.6*29.45/100
M of Ca = 7.5392 grams
Thus, the correct procedure is given by 29.45 / 100 x 25.6 = 7.5392 grams
1) To find the change in enthalpy, determine the difference between the potential energy of the products and the potential energy of the reactants. (on this diagram, C-A) To find the activation energy, find the difference between the potential energy of the reactants and the "peak" of the curve (on this diagram, B-A). For this diagram, both the enthalpy and activation energy are positive.
2) If the reaction was exothermic, enthalpy would be negative, and the potential energy of the reactants would be greater than the potential energy of the products.
