The NaOH is sodium hydroxide, which is a strong base. The mM seems to be a garbled term. Possibly you are thinking of ml, milliliters.
6N NaOH = 6 mol NaOH
in one liter water 6M NaOH is 6 mol/liter
Hi, 6N NaOH = 6M NaOH 6M NaOH are 6 moles in 1L Mw (NaOH) = 39.88 gr/mole so: m = n x MW = 6 x 39.88 = 239.28 gr NaOH. :)
for NaOH normality = molarity ; hence for 1M(1N) solution you have to dissolve 40 g NaOH in 1L water ...... therfore for 0.1 N soln you have to dissolve 4 g in 1 L water and then standerized it with acid which has known normality
8 grams NaOH (1 mole NaOH/39.998 grams) = 0.2 moles NaOH
No, NaOH is a compound
NaOH and HCl
Hi, 6N NaOH = 6M NaOH 6M NaOH are 6 moles in 1L Mw (NaOH) = 39.88 gr/mole so: m = n x MW = 6 x 39.88 = 239.28 gr NaOH. :)
6n+5 = 11
-5=6n+7 -5-7=6n+7-7 -12=6n -12/6=6n/6 -2=n
6n-1 = 5
6n-15 = -9
6n-1248 = -1242
5n + 6n - 2n
6n + 5n is 11n.
6n/10=3 6n=30 n=5
6n+3=15 6n+3-3=15-3=12 6n = 12 n=12/6 = 2
So far, the best and most general pattern found is that, over three, all prime numbers are of the form 6n +/- 1. In other words, they're either 6n - 1 or 6n + 1, for some n. Here is why this is true. We could do a proof by contradiction and assume that all the natural numbers greater than or equal to 5 are prime. (of course they are not!) We start with5 which is 6-1. The numbers would then be 6n - 1, 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4, and 6n + 5 for some natural number n. If it is 6n, then the number is divisible by 6. When it is 6n + 2, the number is the same as 2(3n+1) so it is divisible by 2. Consider 6n + 3, the number is 3(2n+1), so it is divisible by 3. Last look at 6n + 4, the number is divisible by 2, for it's 2(3n + 2). Therefore all numbers of the form 6n, 6n + 2, 6n + 3, and 6n + 4 are not prime. The only possibilities this leaves are 6n - 1 and 6n + 1. This entire thing can be written more elegantly with congruences, but the goal here was simplicity! There are many other patterns in primes. See the attached link to see them.
15/6n + 6/6n = 21/6n = 7/2n