The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.
We must establish the fact that energy is neither created nor destroyed.
Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.
Total heat absorbed by water, H(water) is;
Q(water) = m C (T2 - T1)
Q(water) = 500 × 4.184 × (100-20)
Q(water) = 167.36 kJ
Consequently, the quantity of heat evolved by the combustion of glycerol is;
Q(glycerol) = 167,360 J = n × ΔH°comb
where, n = no. of moles of glycerol.
167.36 kJ= n × 1654 kJ/mole
n = 167.36/1654
n = 0.1012 moles of glycerol.
Therefore, mass of glycerol combusted, m is;
m = n × Molar mass
m = 0.1012 × 92.09
m = 9.32 g.
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Answer:
Option E. Zirconium
Explanation:
From the question given above, the following data were obtained:
Length of side (L) of cube = 0.2 cm
Mass (m) of cube = 52 mg
Name of the unknown metal =?
Next, we shall determine the volume of the cube. This can be obtained as follow:
Length of side (L) of cube = 0.2 cm
Volume (V) of the cube =?
V = L³
V = 0.2³
V = 0.008 cm³
Next, we shall convert 52 mg to g. This can be obtained as follow:
1000 mg = 1 g
Therefore,
52 mg = 52 mg × 1 g / 1000 mg
52 mg = 0.052 g
Thus, 52 mg is equivalent to 0.052 g.
Next, we shall determine the density of the unknown metal. This can be obtained as follow:
Mass = 0.052 g.
Volume = 0.008 cm³
Density =?
Density = mass / volume
Density = 0.052 / 0.008
Density of the unknown metal = 6.5 g/cm³
Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium
Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
Explanation:
the answer and explanation is in the picture
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