Mass of water because then it will change his States of matter
I think that you have put up an incomplete question. However, i am answering the question based on my research and knowledge.
Lissa- accuracy and precision are both low
Lamont- accuracy and precision are definitely high
<span>Leigh Anne- accuracy is low but precision is definitely high.
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I hope that this is the answer that you were looking for and the answer has definitely come to your desired help.
The average atomic mass of the imaginary element : 47.255 amu
<h3>Further explanation </h3>
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% ..
isotope E-47 47.011 amu, 87.34%
isotope E-48 48.008 amu, 6.895
isotope E-49 50.009 amu, 5.77%
The average atomic mass :

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.