Metal ore has other elements in it as well. Also sediment and stone might cover the ore. We don't want to have a phone with sediment on it do we? thus these few reasons are why.
The average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
Given:
ΔH° of xenon difluoride (XeF2) = -105 kJ/mol
ΔH° of xenon tetrafluoride (XeF4)= -284 kJ/mol
ΔH° of xenon hexafluoride (XeF6) = -402 kJ/mol
The bond energy of Xe-F in XeF2 can be calculated as follows,
As we know that
ΔH° = ΔH°(bond formed) + ΔH°(bond broken)
The chemical reaction for the formation of XeF2 can be written in such a way,
Xe (g) + F2 (g) → XeF2 (g)
= [1 mol F2 (159 kJ/mol)] + [2(-Xe-F)] - 105 kJ/mol
= 159 kJ/mol + 2(-Xe-F) - 264 kJ/mol
= 2(-Xe-F)
Xe-F = 132 kJ/mol
Thus, we concluded that the average bond energy of the Xe¬F bonds in each fluoride is 132kJ/mol.
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The volume measured using such a cylinder will be reported to the nearest 10th mL.
<h3>Cylinder graduation</h3>
10 mL graduated cylinders are always read to the nearest two decimal places.
100 mL graduated cylinders are always read to the nearest 1 decimal place. The nearest 1 decimal place is the same thing as the nearest 10th.
Thus, a reading made using a 100mL increment graduated cylinder would be reported to the nearest 10th mL.
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Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.
<h3>How is Sulphur hexafluoride formed?</h3>
Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.
Thus, a sample of fluorine yields 7g of SF6.
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