While staying in the same period, if we move from left to right across the period, the atomic radius decreases. The reason is, in a period the number of shells remain the same and the number of electrons and protons increase as we move across the period to the right. The increased electrons and protons attract each other with greater force and hence the atomic size decreases.

So the element on the left most will have the largest atomic radius. So the correct ans is Potassium. Potassium will have the largest atomic size among Potassium, Calcium and Scandium.

8 0

3 years ago

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

2 years ago

The agent of mechanical weathering in which rock is worn away by the grinding action of other rock particles is call

Igoryamba

Physical weathering

3 0

3 years ago

BigorU [14]

Answer:

Samarium

Explanation:

The element Sm describe is called Samarium. This element has unique sets of properties that makes it very unique and distinct.

The lanthanides are found in the f-block on the periodic table of elements.

This element is a moderately hard silvery metal that readily oxidizes in air. It assumes an oxidation state of +3. The element has an atomic number of 62

7 0

3 years ago