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 oxidation states of the elements before and after the reaction is;
Pb oxidation state changes from 0 to +2
SO₄²⁻ ion there's no change in the oxidation state during the reaction
Au oxidation state changes from +3 to 0
reduction reactions are when there's a decrease in the oxidation state of the species
oxidation reactions are when theres an increase in the oxidation state of the species
the element where there's a decrease in oxidation state is Au.
Therefore Au gets reduced.
answer is B) Au
Answer:
0
Explanation:
Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.
Answer:10,0000 years
Explanation:
If it takes 2000 years to weather 1cm of limestone
It will take 5*2000 years to weather 5cm of limestone. This gives us 10000 years.
