Chemistry Forum

Yes, I know, but how can I calculate it with the thermodynamic data provided in the text?
Whether or not the correct result is among those proposed.

Hi everyone, I have this exercise: 'Which of the following values represents the normal boiling point of carbon tetrachloride? Use the thermodynamic data given. CCl4.JPG a)-272°C b) 25°C c) 67°C d) 69°C e) 109°C. " I thought of applying the following formula: ΔG°= 0 ; ΔG°=ΔH°-TΔS° but my result...

Wow thanks

1) If ΔG°_r<0 for a reaction at all temperature, then ΔH°_r is___ and ΔS°_r is____ a) negative, positive. b) positive, negative. c) negative, negative. d) positive, positive. e) positive, either positive or negative. ___________________________________________________________________________________...

I had seen my professor solve a similar exercise like this, but I made a mess, sorry

I made the conversion of the heat capacity of the water from kcal to kJ in a single step, to multiply it with the mass of the water expressed in kg. Since I had expressed |Q|in kJ and being able to make the quotient between two measures in kJ

Hello everyone, here I am again :wink: I have this exercise: "A feverish student weighing 75 kg is immersed in 400 kg of water at 4.0°C to try to lower the fever. The body temperature drops from 40.0°C to 37.0°C. Considering that the specific heat of the student's body is 3.77 J/(g *°C), what i...

Great, thanks so much

ah ok, I understand what you were asking me
the total moles I calculated, adding n(NaOH) and n(HCl), are 0.349 mol
but they are not in a 1: 1 ratio, with these data NaOh is the limiting reagent.
Q=-ΔH°r (heat released)
Q=-ΔH°r*n= -(-56.2 kJ/mol)*0.13 mol= 7.306 kJ
it's correct?

what does exercise require?

NaOH + HCl => NaCl + H2O

1mol + 1 mol => 1 mol + 1 mol

right?

Hello everyone, I have this exercise: "The neutralization heat of HCl by NaOH is ΔH°r = -56.2 kJ / mol. How much heat is released when 125 mL of 1,750 M HCl is mixed with 195 mL of 0.667 M NaOH?" Would you help me figure out how to fix it? First I determined n n(NaOH)=M*V= 0.667 mol/L*0.19...

Right, what a fool I was !! Thanks so much

Hi everyone, I tried to do this exercise but I don't get the expected result from my text, would you help me understand where my mistake is? Thank you all. The text of the exercise says: "If you burn at 25 ° C and 1 atm 0.100 dm ^ 3 of ethane, C2H6 (g), measured at 20 ° C and 1.50 atm you get 9...

Hello, sorry if I answer you late, but there was a fire emergency, here the woods have been burning for days. Here is my development: P(CH4)=1.45*10^5 Pa = 1.43 atm n(CH4)= (1.43 atm*2.50 L)/(0.08206*T) = 43.6/T mol P(O2)= 2.7.*10^5 Pa = 2.66 atm n(O2) = (2.66 atm*4.5 L)/(0.08206*T) = 146.1/T mol n ...