Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
I believe it is D. Hope this helps you!
Answer:
reproduction
Explanation:
reproduction, process by which organisms replicate themselves
Answer:
Kp = 0.81666
Explanation:
Pressure of PCl₅ = 0.500 atm
Considering the ICE table for the equilibrium as:
PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)
t = o 0.500
t = eq -x x x
---------------------------------------------
--------------------------
Moles at eq: 0.500-x x x
Given the pressure of PCl₅ at equilibrium = 0.150 atm
Thus, 0.500 - x = 0.150
x = 0.350 atm
The expression for the equilibrium constant is:
So,
x = 0.350 atm
Thus,
<u>Thus, Kp = 0.81666</u>
