C
A has 2H 2O on the left and 4H 2O on the right
B has 4H 4O on the left and 4H 2O on the right
D has 2H 2O on the left and 2H 4O on the right
C has 2H 2O and 2H 2O so is balanced.
The first shell can only hold 2 electrons, but the pupil placed in 8 electrons.
The second shell can hold up to 8 electrons, but the pupil only placed in 2 electrons.
Flourine only has 9 total electrons, yet there are 10 electrons in the diagram.
Answer:
it is not correct I have an answer for ''why''
Explanation:
You need to balance it
s= 2
0=2
S+30•2 --------> SO•3
2 2 2 2
S +3O•2-------->SO•2......MULTIPLY
2 4 2 4......LCM IS 4
2 1 2 1....DIVIDE BY 4
SO NOW WE GOT THE BALANCE NO. WE CAN PUT IT IN THE EQUATIONS.
2S+3O•2 -----> 2SO•3
now you can check the both sides its balances the correct answer is this
POH = -log[OH-] and pH + pOH = 14
So, all you have to do is take the negative log of 2.3*10^-5 and then subtract that number from 14 to find the pH
It is an Alkene because it has a double bond, so it’ll have “ene” at the end. The simplest Alkene has 2 carbons.
2 carbons = “eth”
Look at that! Two carbons! It must be “ethene”