Answer:
I dont know the first blank but I know the second one is groups and the last one is periods.
Answer:
Explanation:
In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case 
) and the anion is the atom at the right of the name (in this case 
). With this in mind, the <u>formula would be</u> 
.
Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> 
. So, we have as reagents:
The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:
With this in mind, the reaction would be:
I hope it helps!
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: No.
Explanation: One mole of zinc is not the same as one atom of zinc. In one mole of zinc, there are approximately 6.022*10^23 atoms of zinc.
