Problem One (left)
This is just a straight mc deltaT question
<em><u>Givens</u></em>
m = 535 grams
c = 0.486 J/gm
tf = 50
ti = 1230
Formula
E = m * c * (ti - tf)
Solution
E = 535 * 0.486 * ( 1230 - 50)
E = 535 * 0.486 * (1180)
E = 301077
Answer: A
Problem Two
This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.
E = H m
H = 2257 J/gram
m = 11.2 grams
E = 2257 * 11.2
E = 25278 to three digits is 25300 Joules. Anyway it is the last one.
Three
D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.
Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.
Answer: B
Four
The best way to think about this question is just to get the answer. Answer C.
A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.
B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.
C: is correct because the molecules neither stop nor do they stick. The hit and move on.
D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.
E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.
Five
Half life number 1 would leave 0.5 grams behind.
Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.
Answer: 0.25
Answer C
Answer :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule. The electrons are represented by dot.
The given molecule is, perbromate ion.
Bromine has '7' valence electrons and oxygen has '6' valence electron.
Therefore, the total number of valence electrons in perbromate ion,
= 7 + 4(6) + 1 = 32
According to Lewis-dot structure, there are 14 number of bonding electrons and 18 number of non-bonding electrons.
Formula for formal charge :

The Lewis-dot structure of perbromate ion is shown below.
Answer:
a. Ca₃(PO₄)₂.
b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced
c. 3.1g of Ca₃(PO₄)₂
d. Percent yield = 93.5%
Explanation:
a. Based on the reaction:
3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)
<em>3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.</em>
<em />
As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,
<h3>Ca₃(PO₄)₂</h3><h3 />
b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate
0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =
<h3>0.010 moles of Ca₃(PO₄)₂ can we expect to be produced</h3><h3 />
c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;
0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =
<h3>3.1g of Ca₃(PO₄)₂</h3><h3 />
d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

<h3>Percent yield = 93.5%</h3>
Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher levels Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular color. As a result of all these jumps, a spectrum of colored lines will be produced. The color you see will be a combination of all these individual colors.