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3 years ago

5

How to first ionization energies vary within groups and across periods

1 answer:

trapecia [35]3 years ago

8 0

Answer:

Ionisation energy increases along a period but decreases down a group

Explanation:

As you go along the period, the amount of protons in the nuclear increases. This causes nuclear attraction to increase meaning the attraction between the protons and the electrons is higher. This causes the electrons to be held more tightly to the nucleus meaning they are harder to remove. As well as that, the shielding stays the same along a period since the electrons are being added to the same shell, just different orbitals. Overall this makes it harder to remove an electron as you go along a period causing the ionisation energy to increase along a period.

On the other hand when you go down a group, the atomic radius increases. This is because the electrons are added to new shells which are further away from the nucleus, this also causes shielding to increase as there are more shells. This means the outer electrons being removed are held less tightly by the protons in the nucleus meaning they are easier to remove. This means that the ionisation energy decrease down a group.

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Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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3 years ago

A frying pan needs a Teflon coating of 1.00 mm in thickness to cover an area of 36.0 square inches. How many ounces of Teflon ar

Otrada [13]

Answer:

m = 0.659 ounce

Explanation:

It is given that,

The thickness of a Teflon coating is, d = 1 mm

Area of the coating, A = 36 inch²

The density of Teflon, d = 0.805 g/mL

We need to find ounces of Teflon are needed.

Firstly, find the volume of the Teflon needed,

1 inch² = 6.4516 cm²

36 inch² = 232.258 cm²

Density,

$\rho=\dfrac{m}{V}$

V is volume of the Teflon needed, V = Ad

So,

$m=\rho V\\\\m=0.805\ g/cm^3\times 232.258\ cm^2 \times 0.1\ cm\\\\m=18.69\ g$

Also, 1 gram = 0.035274 ounce

18.69 gram = 0.659 ounce

So, 0.659 ounces of Teflon are needed.

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Tomtit [17]

Answer:

solid

Explanation:

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irinina [24]

Answer:

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astraxan [27]

Answer:

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The water molecules are only weakly attracted to the wax, so the cohesive forces pull the water molecules together to form beads .

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