<em><u>the</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>aluminium</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>1</u></em><em><u>4</u></em>
Answer:

Explanation:
1. Moles of CCl₄

2. Molar mass of CCl₄
MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol
3. Mass of CCl₄

4. Volume of CCl₄

Answer:
B
Explanation:
Look on the x-axis for the tick marked "60". This indicates 60 degrees Celsius, which we want. Now, look on the y-axis for the tick marked "60". This indicates 60 grams of
. Trace along the graph to find where these two places meet at (60, 60).
Now, look for the solubility curve of
; it's the yellow-orange line. Find out what the y-coordinate of the point where x = 60 is on the line: it's around (60, 65).
So, since the point (60, 60) is below the line corresponding to this substance,
is unsaturated.
The answer is B.
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62