1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^4
(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6
Answer:
Answer of question a is 345J.
Explanation:
In question a following is given in data:
-mass of iron (m) = 10.0 g
-temperature (ΔT) = final temperature- initial temperature= 100-25= 75 degree Celsius
-Specific Heat capacity of iron (c)= 0.46J/g°C.
Heat (Q)=?
Solution:
Formula for Heat is :
Q=m x c x ΔT
Q= 10 x 0.46 x 75
Q= 345 J.
so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.
- From the above formula all other questions can easily be solved from the same procedure.
The noble gases are relatively unreactive because they have a stable octet of valence electrons.
Thus, they do not tend to undergo reactions in which they will gain or lose valence electrons,
However, <em>only He, Ne, and Ar are inert</em>. Kr and Xe combine with other highly reactive elements to form stable compounds.