B: nothing the color of the solution when heated
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.
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
Isotope ¹⁸F⁻ contains:
1) p⁺ = 9; number of protons.
Fluorine has a<span>tomic number Z = 9 (total number of protons).
2) e</span>⁻<span> = 10; </span>number of electrons.<span>
In element number of electrons and protons are the same, because element has neutral charge, but because in this example, fluorine is anion with negative charge, it has one electron more.
3) n</span>° = 9; number of neutrons.
<span>Mass number
A = 18 is total number of protons and neutrons in a nucleus, so number of neutrons is A-Z = 18-9=9.</span>
The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.
