posted by alex on .
A 100.0 mL portion of 0.250 M calium nitrate solution is mixed with 400.0 mL of .100M nitric acid solution. What is the final concentration of the nitrate ion?
BOB PURSLEY PLEASE EXPLAIN MORE THOROUGHLY HOW YOU GOT THE NITRATE ION TO BE .5M
Well, you are right. The Ca(NO3)2 IS 0.25 M just as you said; however, the problem asks for NO3^- and since there are two moles NO3^- per 1 mole Ca(NO3)2, that makes the NO3^- just twice or 0.250 x 2 = 0.5 M and you took 100 mL of that so M x L = moles = 0.05 moles just like Bob P wrote.
thank doctor bob pursley that's right i know who you are.
No. I can guarantee you that Bob P and DrBob222 are NOT the same person. In fact, he and I live about 400 miles from each other. I just picked up your confusion over the nitrate ion because he wasn't on-line at the moment.
o oops hahah thankss for answering my question!
1)How many milliliters of 1.5 M NaOH will react completely with 448 mL of 0.600 M HCl?
NaOH (aq) + HCl (aq) ¨ NaCl (aq) + H2O (l)
2)What is the percent yield of the following reaction if 15 g H2O are obtained from the combustion of 18 g C2H6?
2 C2H6 (g) + 7 O2 (g) ¨ 4 CO2 (g) + 6 H2O (g)
3)What is the molarity of a solution in which 6.4 g HCl is dissolved in enough water to make 2.4 L solution?
4)How many atoms of H are there in 4.3 g CH4?
I REALLY NEED HELP SOLVING THESE CHEM PROBLEMS