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.
Try adding spaces next time! That's iodine. Check all of the numbers to make sure all of the orbitals are filled, then find the ones which aren't. In this one, only the 5p5 subshell isn't full. 5p5 is the fifth row on the right side, count across the nonmetals and metalloids until the fifth one (a halogen). That's iodine, and that's your answer!
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.
An ion can be cation (positively charged) or anion (negatively charged).
Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).
The remaining elements are completed as follows to attain octet structure;
<u>Element</u>--<u>valence electron</u>--<u>electrons to gain</u>--<u>electrons to lose</u>--<u>ion formed</u>
O ------------ 6 ---------------------- 2 ------------------------ none -------------- 
Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ 
Br ----------- 7 --------------------- 1 ------------------------ none --------------- 
S ------------ 6 ----------------------- 2 ------------------------ none --------------- 
Cl ------------ 7 ----------------------- 1 ------------------------ none ----------------
K -------------- 1 ----------------------- none ----------------------- 1 ------------------ 
Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- 
Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- 
Learn more here: brainly.com/question/21089350