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Marrrta [24]
3 years ago
13

A metal (M) forms an oxide with the formula MO. If the oxide contains 7.17 % O by mass, what is the identity of the metal?

Chemistry
1 answer:
Nataly_w [17]3 years ago
7 0

The metal is lead, Pb.

One unit of the oxide contains one atom of O (16.00 u).

∴ Mass of oxide = 16.00 u O × (100 u MO/7.17 u O) = 223.15 u MO

Mass of M = mass of MO – mass of O = 223.15 u -16.00 u = 207.2 u

The only element with an atomic mass of 207.2 u is lead (Pb) and the formula of the oxide is PbO.

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Adding powdered limestone (calcium carbonate) to a lake affected by acid rain can decrease the availability of phosphate ion, an
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5 0
4 years ago
Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn=
IgorC [24]

Answer: \Delta H^0=+0.3kJ/mol.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

S_{rhombic}+O_2(g)\rightarrow SO_2(g)    \Delta H^0_1=-296.06kJ   (1)

S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ  (2)

The final reaction is:  

S_{rhombic}\rightarrow S_{monoclinic}  \Delta H^0_3=?   (3)

By subtracting (1) and (2)

\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ

3 0
3 years ago
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