X is correct to a point. There are 85*2.6 = 221 g NaNO3. BUT there are NOT 14 g N in that. %N in NaNO3 is (14/85)*100 = about 16.47% so in 221 g NaNO3 there are 221*0.1647 = about 36.40 g N or 36.40/14 = 2.60 mols N. Of course that's the long way to go. You can see 1 N ato...
There is 1 mol N/1 mol NaNO3; therefore there must be .....mole N in 2.6 mol of NaNO3.
How many mols do you need? That's mols K2Cr2O7 = M x L = ? Then mols = g/molar mass. You know mols ad molar mass. solve for grams.
__2K + __CaBr2 → __2KBr + __Ca
Use PV = nRT with R 8.314 and P comes out directly in kPa OR use R = 0.08206 in L*atm/mol*K and find P in atm, then atm x 101.325 = kPa.
C3H8 + 5O2 3CO2 + 4H2O mols C3H8 = grams/molar mass Using the coefficients in the balanced equation, convert mols C2H8 to mols CO2. Now convert mols CO2 to L remembering that 1 mol occupies 22.4 L OR you can use PV = nRT and solve for V in L../
2NaOH + 2Al + 6H2O 2NaAl(OH)4 + 3H2 Use PV = nRT and solve for n = mols H2 OR remember at STP 1 mol occupies 22.4L. Using the coefficients in the balanced equation, convert mols H2 to mols Al Then mols Al = grams/atomic mass. You know atomic mass and mols, solve for g...
answered above by jai
mols CH3COOH (HAc) = M x L = 0.581 mols CH3COONa (NaAc) = 0.581 .......HAc + OH^- ==> Ac^- + H2O I......0.581..0......0.581........ add.........2*0.054............... C.....-1.08..-1.08..+1.08 E.....0.499..0.......1.66 Substitute the E line into the Henderson-Hasselbalch eq...
Hi is an acid; KOH is a base. They partially neutralize each other. HI + KOH ==> KI + H2O You are right. Find the OH^-. mols HI = M x L = about 0.495 mols KOH = M x L = ? Subtract to find the one in excess and M = mols excess/total volume in L. Then pH = -log(H^+) if HI in ...