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umka2103 [35]
4 years ago
14

In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element

s has the smallest atomic size ?
Chemistry
1 answer:
wariber [46]4 years ago
6 0
Answer:
            Phosphorous has the smallest atomic size.

Explanation:
                   As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

<span><span>Magnesium          0.160 nm
</span><span>
Aluminium           0.130 nm
</span><span>
Silicon                  0.118 nm
</span><span>
Phosphorus         <span>0.110 nm</span></span></span>
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Diamond is an allotrope of :
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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
What are the coordinates of the midpoint of the line segment with endpoints (2, 5) and (4, -7)?
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