Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
The second one, vocab and definition doesn't add up
Characteristic properties are used because the sample size and the shape of the substance does not matter.
Explanation:
Answer
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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0
2× ( Oxidation number of Al) +3(−2)=0
2× ( oxidation number of Al) +6
∴ Oxidation number of Al =+3
Electronic configuration of cromium is
Cr-[Ar]4s¹3d⁵
When cromium loses two electrons it becomes Cr⁺².
So its electronic configuration becomes,
Cr⁺²-[Ar]3d⁴
One electron will go from 4s orbital and one electron will go from 3d orbital.
So the answer here is D. [Ar]3d⁴ -because after losing 2 electrons electronic configuration of cromium becomes [Ar] 3d⁴.