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:
4.52×10^24
Explanation:
N = n × Na
where; N = no. of bananas
n = no. of moles
Na = Avogadro's constant
Which is 6.02×10^23
N = 7.5 × 6.02×10^23
N =4.515×10^24
Answer:
The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. ... This means that they have more collisions with each other and the sides of the container and hence the pressure is increased.
Answer: weigh is m = n × M = 2.87 mol × 58.44 g/mol
Explanation: mass = amount of substance × molar mass
M((NaCl) = 22.99 +35.45