question 1: a mixture of 3 gases exerta atotal pressure of 1.5atm. the mixture c
ID: 683929 • Letter: Q
Question
question 1: a mixture of 3 gases exerta atotal pressure of 1.5atm. the mixture contains .31moles of ch4,0.25moles of c2h6...how many moles of the third gas c3h8 arepresent?question 2:
what will the pressure exerted by 10 grams of N2 in a 3L containerat a tempeature of 30*c?
question 3:
lithium hydroxide reacts with CO2 as follows:
2LiOH + CO2-->Li2CO3 + H2O
how many grams of LiOH will react with 3L of CO2 at 25*C and1atm?
question 4:
what was the initial temp. of a 40gram block of aluminum(C=0.903J/G*C) which lost 2kJ of heat while cooling to 25*C?
question 5:
how much heat is released in the formation of 10L of nh3 at 25*C,1atm given:
2N2+ 3H2-->2NH3 (delta H for the rxn is -92.6kJ)
and last
question 6:
calculate delta H for the following rxn:
2H2S + 3O2-->2H2O + 2SO2 question 1: a mixture of 3 gases exerta atotal pressure of 1.5atm. the mixture contains .31moles of ch4,0.25moles of c2h6...how many moles of the third gas c3h8 arepresent?
question 2:
what will the pressure exerted by 10 grams of N2 in a 3L containerat a tempeature of 30*c?
question 3:
lithium hydroxide reacts with CO2 as follows:
2LiOH + CO2-->Li2CO3 + H2O
how many grams of LiOH will react with 3L of CO2 at 25*C and1atm?
question 4:
what was the initial temp. of a 40gram block of aluminum(C=0.903J/G*C) which lost 2kJ of heat while cooling to 25*C?
question 5:
how much heat is released in the formation of 10L of nh3 at 25*C,1atm given:
2N2+ 3H2-->2NH3 (delta H for the rxn is -92.6kJ)
and last
question 6:
calculate delta H for the following rxn:
2H2S + 3O2-->2H2O + 2SO2
Explanation / Answer
I will help you, but next time if you want help fast don'tpost more than one question at a time. If you have sixquestion make six posts, and you will get faster answers. question 1: Use PV=nRT use standard volume and temperature for the equation. torelate moles and temp. P=1.5 atm V= 1 L n=.31+.25+x R=.0821 T=293 K Solve for the x. x=.5 mol Question 2 Same equation PV=nRT sovle for P V=3L mol N2=10g/28.02g/mol=.357mol R=.0821 T=30+273=303K P=2.96atm question 3 find the number of mol of CO2 used and convert to grams 1atm * 3L=n*.0821*298 n=.1226mol .1226*2=.2452 mol LiOH used .2452mol*23.95g/mol=5.78g reacted question 4 just find the temperature difference by this equation 2000J/.903J/gC=2214g*C 2214g*C/40g=55.4 C add that to the original temp 55.4+25=80.4 C original temp question 5 solve for the number of moles formed.divide the delta H by twoto get the kJ/mol and multipy 10L*1atm=n*.0821*298K n=.408 mol -92.6/2=-46.3 kJ/mol -46.3*.408mol=-18.9 kJ energy released question 6 I wil leave to you. You just need to look upthe heats of formation for SO2 and H2O then substract that from theheat of formation H2S and times the final answer by two because thereaction uses two moles and forms two moles. You don't needthe number for O2 becaue the heat of formation for that is0.Related Questions
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