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r-ruslan [8.4K]
3 years ago
6

Molar mass of Fe2O3 = g/mol

Chemistry
1 answer:
AfilCa [17]3 years ago
3 0

Answer:

159.6882 grams

Explanation:

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Generally, when going down a group on the periodic table:
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Generally when we move down the group on a periodic table the atomic radii increases as the valency electrons occupy higher levels due to the increasing quantum number. Hence the atomic radii increases down the group.

The ionic radii increases down the group because while we move down the group the elements gain electrons and form ions called anions as an additional electron occupies the orbital the ions get bigger in size. Hence the ionic radii increase.

Electronegativity is described as the ability to attract and bind with electrons and it is a qualitative property. It decreases as we move down the group because the distance between the valency electrons and the nucleus increases. Hence electronegativity decreases down the group.

Reactivity increases as we move down the group as the metals have the tendency to lose electron form its outer shell.

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The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
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When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

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The replacing of sodium hydroxide with potassium hydroxide (KOH) to the reaction will least affect the organic product that forms.

 

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