heat=masswater*c*changetemp
= 50ml*1.02g/ml*cwater*(36.7-23.8)
look up c water, then after that, somehow you need to figure how many moles of HCl you had.
Information:
Original temp of HCl: 23.3C
of NaOH: 23.8C
Final temp of the mixture: 36.7C
I am to find qH2O (assume 50 mL of solution and use average density of 1.02 g/mL) but I'm a bit lost.
I also need to find /\H and /\H per mole of H+a and OH- ions reacting, but I assumed I could not answer these 2 without qh20
4 answers
Is that 50 mL total? How much HCl and how much NaOH? What's the molarity of HCl and NaOH? You need to know mols H2O formed if you are to calculate dH/mol.
I see Bob Pursley ignored the difference in temperature. Is that 23.3 and 23.8 a typo?
So 50*1.02*4.18*12.9=2750J for qH2O?
25 mL of each solution. The molarity for each is 2 M.
Molarity is moles per liter, yes? So if I multiply the molarity, 2, by .025 L ... the moles of both will be .05? Or am I totally off?
25 mL of each solution. The molarity for each is 2 M.
Molarity is moles per liter, yes? So if I multiply the molarity, 2, by .025 L ... the moles of both will be .05? Or am I totally off?