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.
The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
<h3>Weight of one gallon of water</h3>
The weight of 1 gal of water is given as 3785 g
Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g
<h3>Volume of the water in cubic meters</h3>
Volume = mass/density
Volume = 3.2 x 10¹² g/1 gmL
Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³
Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
Learn more about volume here: brainly.com/question/1972490
#SPJ1
Lithium Bromine (LiBr) is formed from them....
The answer is C2O.............
562 grams because mass can not be created or destroyed