23) A fixed amount of gas at 25.0 C occupies a volume of 10.0L when the pressure
ID: 580962 • Letter: 2
Question
23) A fixed amount of gas at 25.0 C occupies a volume of 10.0L when the pressure is 751 torr reduced to 7.88 L at a constant Boyle's law to calculate the pressure (torr) temperature of 25.0"C when the volume is E) 5.92 104 D) 953 A) 0.105 B) 1.25 592 the 24) The value of AH for the reaction below is -790 kJ. The enthalpy change reaction of 2.95 g of S is_ kJ. (Molar Mass of S= 32.07 g) 25(s) + 302 (g) 2503() E) 12 A) 23 B) 36 C)-790 D) -12 25) The value of }r for the reaction NaOH is formed in the reaction? below is-126 kJ· -kj are released when 300 moles of 2Na202 (s) + 2H20 (I) 4NOH (s) + (g) A) 7.8 B) 63 D) -126 E) 94.5 C 252 26) Which one of the following is an exothermic process? A) condensation of water vapor B) water evaporating Q) boiling soup D) ice melting E Ammonium thioganatc and barium hydroside are mived at 25C: the semperature drops 27) How many atoms of neon occupy a volume of 14.3 L at STP? A) 1.92 x 1025 atoms B) 2.21 x 1025 atoms ) 3.88 x 1024 atoms D) 3.84 x 1023 atoms E) 9.45 x 1023 atoms 28) Given the following reactions 2A 3B H =-100 kJ the enthalpy of reaction for is kJ. A) 400 B)-350 C) -200 D) 600 E) 150
Explanation / Answer
(23)
Using Boyle's law:
P1V1 = P2V2
Putting values:
751*10 = P2*7.88
Solving we get:
P2 = 953 torr
(24)
Moles of S = Mass/MW = 2.95/32.07 = 0.092
In the reaction, 2 moles of S undergo combustion to produce 790 kJ heat.
So, heat produced when 0.092 moles of S undergo combustion = 0.092*(790/2) = 36 kJ
(25)
In the reaction given, 4 moles NaOH are produced along with 126 kJ of heat.
So, when moles of NaOH are produced, heat released = 126*(3/4) = 94.5 kJ
(26)
Condensation of water vapor is an exothermic process, because in this heat is released when vapor get converted to liquid droplets.
Hope this helps !
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