00:01
The goal of the problem is to determine the chemical formulas of the following compounds.
00:06
Let's first start with ribidium nitrate.
00:10
Rebidium is rb plus and nitrite is no2 minus.
00:16
The next step would be to cross the charges.
00:19
What i mean by this is that the charge of the ribidium would end up being the subscript for the nitrite.
00:26
And the charge of the nitrite, which is a minus, would be the subscript for the ribidium.
00:32
So when we write it, we would write it as rb, n02, since both of them have a one charge, plus and minus.
00:41
Next is potassium sulfide.
00:43
Potassium is k plus, while sulfide is an s -2 -minus.
00:49
So when we cross the charges, the two would be the subscript for k, and s would have a subscript of 1, leading to k -2s.
00:58
Next we have sodium hydrogen sulfide.
01:01
Sodium has a plus 1 charge, hydrogen has a plus 1 charge as well, and sulfide has a 2 minus charge.
01:11
Now since there's 3 ions, we have to balance the charges.
01:15
The n -a and h have a plus plus charge, and the s has a 2 -minus charge.
01:20
Therefore the charges will be balanced, and you just need to write the ions as it's.
01:24
So we write nahs, and that would be our ionic compound.
01:30
Next we have magnesium phosphate.
01:32
Magnesium has a plus 2 charge, and phosphate is p .04, 3 minus.
01:38
So when we cross the charges, m .g would get the 3 minus charge, you would get m .g 3, and p .o4 would get the plus 2 charge.
01:47
So our full compound would be mg3, p .o .4 .2.
01:51
Next we have calcium hydrogen, hydrogen, phosphate.
01:54
Calcium has a plus 2 charge, hydrogen has a plus charge, and phosphate has a 3 minus charge.
02:00
Similar to a previous problem, you just need to add up the charges and balance them.
02:05
So since ca and h both have plus and plus, they add up to 3 plus, which would balance out with the 3 minus...