Answer:
a) Endothermic
b) T₂ = 53.1 ºC
Explanation:
a) We are told that when the ammonium nitrate dissolves in water the pack gets cold so the system is absorbing heat from the surroundings and by definition it is an endothermic process.
b) Recall that the heat, Q, is given by the formula:
Q = mcΔT where m is the mass of water,
c is the specific heat of water, and
ΔT is the change in temperature
We can determine the value for Q since we are given the heat of solution for the ammonium nitrate. From there we can calculate ΔT and finally answer our question.
Molar mass NH₄NO₃ = 80.04 g/mol
moles NH₄NO₃ = 50.0 g/ 80.04 g/mol = 0.62 mol
Q = 25.4 kJ/mol x 0.62 mol = 15.87 kJ = 15.87 kJ x 1000 J = 1.59 x 10⁴ J
Q = mcΔT ⇒ ΔT = Q/mc
ΔT = 1.59 x 10⁴ J/ (135 g x 4.184 J/gºC ) = 28.1 ºC
T₂- T₁ = ΔT ⇒ T₂ = ΔT + T₁ = 28.1 ºC +25.0 ºC = 53.1 ºC
HCl is a monoprotic acid, which means that each mole of HCl releases one mole of hydrogen ions upon dissociation. Therefore, we calculate the moles of HCl present using:
Moles = Molarity * Volume (in liters)
Moles = 11.6 * 0.015
Moles = 0.174 moles of HCl = moles of H+ ions
Now, we use the same formula to calculate the molarity of the new solution, since the number of moles remains constant.
0.174 = M * 0.5
M = 0.348 M
The molarity of the new solution is 0.348
We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)= 
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.
The answer to problem is [He] 2s1. Hope it helps
