Number of protons as stated in the question = 92.
a. 92
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:
4 & 5 have the same number of elements
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01
A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite
It starts at the bottom and then it goes up into the air as evaporation and that’s when the air gets cooler so gas is cooler and the liquid would be hotter
