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3 years ago

Which group/family contains elements with the smallest radius within the same period?

Chemistry

1 answer:

nasty-shy [4]3 years ago

5 0

Answer:

Halogens

Explanation:

From the given choices, the halogens will have the smallest radius within the same period.

The size of an atom is estimated by the atomic radius. This is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state.

Across a period in the periodic table, atomic radii decrease progressively from left to right.
Down a group from top to bottom, atomic radii increase progressively due to the addition of successive shells.

Since halogen is the right most group from the choices given, it will have the smallest radius.

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Answer:

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3 years ago

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically writ

Alex17521 [72]

Answer:

T = 42.08 °C

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

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R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = 45.6 kJ/mol = 45600 J/mol (As 1 kJ = 1000 J)

$k_2=2\times k_1$

$k_1=0.0160s^{-1}$

$T_1=30\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (30 + 273.15) K = 303.15 K

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$\ln \dfrac{k_{1}}{2\times k_{1}} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )$

$\ln \dfrac{1}{2} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )$

$8.314\ln \left(2\right)=-45600\left(\frac{1}{303.15}-\frac{1}{T_2}\right)$

$8.314\ln \left(2\right)=-150.42058+\frac{45600}{T_2}$

$144.65775 =\frac{45600}{T_2}$

$T_2=\frac{45600}{144.65775}$

$T_2=315.23\ K$

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T(K) = T( °C) + 273.15

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315.23 = T( °C) + 273.15

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