Calculations a technician performs for the CSP should be located on the O a) standard operating procedure (SOP). b) master formulation record (MFR). O) compounding record (CR). d) quality assurance
Explanation:
Each orbital can only carry at most 2 electrons wih opposite spins.
The first 3 shells are fully filled, giving 9 orbitals.
There is an extra electron in 4s, which gives another orbital. (even though it is not filled yet)
Hence there are 10 orbitals.
In a neutral ion, the number of protons would be equal to the number of electrons. You get a positive or negative charge when electrons are lost or gained.
Example: 12/6 C1-
There are 6 protons, 6 neutrons, and 7 electrons.
Ca ionises into Ca^2+. Ca^2+ will be attracted to O^2- ions in the water, since opposite charges attract. (Hydrogen in water forms H^+)
Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:


We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

Now we can <u>put the values into the equation</u>:

Now we can <u>solve for the initial temperature of gold</u>, so:

ºC
I hope it helps!