College Chemistry
posted by Anonymous on .
how do i calculate the moles of Na2CO3 present in the original 25.00mL aliquot of carbonate solution??
the tared mass of Na2CO3 is 1.5179g

the calciul content of person's urine was determined on two different days to be:
day1:
Average Ca(mg/L)=238
Number of measurements=4
day2
Average Ca:268
Number of measurements=5
The analytical method was used known to have a standard deviation of 14mg/L
based on many trials.are the two average above significantly different at 95% confidence level
I used t student test and ended up with t=0.1064
am I right?
thanks 
You don't provide enough information. I know you posted this recently and I told you the same thing. Assuming you have more data, you can use one of the following.
mols Na2CO3 = M x L if you know those values.
mols = grams/molar mass if you know those values. 
I may be reading the problem wrong but I think the problem is asking you to compare the two sets of values and see if there is a statistical difference at the 95% confidence level. Equations are available for that. The one I have is
X1X2 = +/ ts_{pooled}*sqrt(N1+N2/N1N2) and for t you pick that from the table using n1+n22 degrees of freedom. X1 and X2 are the two sets of AVERAGES. 
) At a certain temperature a 2.00 L flask initially contained only 0.600 mol PCl3(g) and 0.0175 mol PCl5(g). After the system had reached equilibrium, 0.00400 mol Cl2(g) was found in the flask. Gaseous PCl5(g) decomposes according to the equation
PCl5(g) ↔ PCl3(g) + Cl2(g)
Calculate the equilibrium concentration of all species and the value of K.