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Illusion [34]
3 years ago
13

Compared to a neon atom, a helium atom has a smaller atomic radius

Chemistry
2 answers:
Brums [2.3K]3 years ago
8 0

Answer:

True

Explanation:

Atomic radius is the distance between the central nucleus of an atom to the outermost shell (containing electrons). Generally, atomic radius increases down a group and the situation in the question is no exception. The increase down a group is because of the increase in electron shell down a group.

Helium (He) has just one electron shell while Neon (Ne) has two electron shells. The distance between the nucleus and the second electron shell of neon is higher than the distance between the nucleus and the single electron shell of helium.

tamaranim1 [39]3 years ago
3 0

Answer:

True

Explanation:

It is true

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Which particle changes to create an isotope of an atom? <br> A.proton<br> B.neutron<br> C.electron
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A

Explanation:

proton

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In general, non-spontaneous reactions __________.
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What is the mass of a sample of a sample of iron (cp=0.44j/g•°C) when it is heated from 60°to 160°C. The heat is transferred is
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5 0
3 years ago
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
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