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
Group 17
Please brainliesttt
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:
We make a simplification because x<<< 0.0225:
x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
Physical change because ther is force which is the electric current
