Answer:
Explanation:
B)The compound potassium chloride (KCl) has different properties than potassium (K) and chlorine (Cl).
Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
Explanation:
Building Vocabulary
Match each term with its definition by writing the letter of the correct definition on
the line beside the term in the left column.
5. nucleus b
6. proton f
7. neutron h
8. electron d
9. atomic number g
10. isotopes c
11. mass number a
12. energy level e
a. the sum of protons and neutrons in the nucleus of an
atom
b. the very small center core of an atom
c. atoms of the same element that differ in the number
of neutrons, but have the same number of protons
d. the particle of an atom that moves rapidly in the
space outside the nucleus
e. a specific amount of energy related to the movement
of electrons in atoms
f. the particle of an atom with a positive charge
g. the number of protons in the nucleus of every atom
of an element
h. the particle of an atom that is neutral
-. mass number a.
12. energy level e
To determine whether an element is paramagnetic or diamagnetic, you need to examine the electron configuration of the element. If it has unpaired electrons, then the substance is paramagnetic. If the electrons are paired, then it is diamagnetic.
