pem

A .600 g sample of CaCO3 is disolve in concentrated HCl (12M). the solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant. 1 What is the molarity of the Ca2+ (calcium ion) in the flask? 2 How many mole...

you are a 13

13 *2 = 26 (more than 20 and less than 40) 13 *4 = 52 (no restriction) 13 *6 = 78 (no restriction)

yes the calcium carbonate has (MW= 100.09 g)hence the moles for CaCo3 is correct but how do you calculate the mole for calcium ion Ca2+

2HCl+CaCO3=CaCl2 + H2O + CO2... i think so....

5d(d^2 - 16) d^2 = 16 (then you take the square root therefore d = 0, -4, 4

Zn(s) + HCl(aqu)^ heat = ZnCl (zinc chloride)+ H(gaseous)

-.4,-.41%,1/5, .44, 404.......i think so

isn't the percentile a -ive #?

try (2x+5)(2x+5)

5/7= .71 draw five pies....each person will receive a little bit over 1/2 of a whole pie

isn't it an imaginary number