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2 years ago

Which element has the smallest atomic radius? calcium potassium scandium titanium

Chemistry

2 answers:

Alex787 [66]2 years ago

8 0

The element that has the smallest atomic radius is titanium

explanation

Calcium,potassium, scandium and titanium are in period 4 in the periodic table.

Generally the atomic radius decreases across the period (from the left to the right) . This is because across the period there is increase in nuclear charge which make electrons been strongly attracted to the nuclear.

Titanium is in atomic number 22 thus it has more nuclear charge that calcium,potassium and scandium therefore titanium has the smallest atomic radius.

nata0808 [166]2 years ago

8 0

Answer: Option (d) is the correct answer.

Explanation:

It is known that in a periodic table, atomic size decreases when we move from left to right.

The given elements potassium, calcium, scandium, and titanium all belongs to third period.

Therefore, out of these elements potassium is at the extreme left and titanium is towards the right. Thus, we can conclude that titanium is the element which has the smallest atomic radius.

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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer: The rate constant at 324°C is $61.29M^{-1}s^{-1}$

Explanation:

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$

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Answer:

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Explanation:

Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.

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37.5 = vf - 50

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