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.
Answer:
+125.4 KJmol-1
Explanation:
∆H C4H10(g) = -2877.6kJ/mol
∆H C(s)=-393.5kJ/mol
∆H H2(g) = -285.8
∆H reaction= ∆Hproducts - ∆H reactants
∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]
∆H reaction= +125.4 KJmol-1
Hey there!:
K = Ka * Kb / Kw
Ka = 1.8*10⁻⁴
Kb = 10⁻¹⁴ / 6.8*10⁻⁴
K = 1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )
K = = 1.8 / 6.8
K = 0.265
Answer A
Therefore:
K is less than on the forward reaction is not favorable .
Hope That helps!
Answer:
.....false
Explanation:
the particles in the air are packed apart whereas the particles in steel are closely packed and sound travel faster in solids than in gasses..
Answer:
The answer to your question is 0.113 moles of Fe₂O₃
Explanation:
Data
moles of Fe₂O₃ = ?
mass of Fe₂O₃ = 18 grams
Process
1.- Calculate the molar mass of Fe₂O₃
Fe₂O₃ = (56 x 2) + (16 x 3)
= 112 + 48
= 160 g
2.- Use proportions to solve this problem. The molar mass is equivalent to 1 mol.
160 g of Fe₂O₃ --------------- 1 mol
18 g of Fe₂O₃ ---------------- x
x = (18 x 1)/160
x = 0.113 moles of Fe₂O₃
