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NNADVOKAT [17]
3 years ago
7

The symbol for xenon xe would be part of the noble gas notation for the element

Chemistry
2 answers:
Allushta [10]3 years ago
8 0
The symbol for xenon xe would be part of the noble gas notation for the element of radium, it is a chemical element with the symbol of Ra and the atomic number of radium is 88. It is the sixth element in group 2 of the periodic table, also known as the alkaline earth metals.<span> </span>
Mademuasel [1]3 years ago
4 0

Answer : The noble gas notation is often given to the short or abbreviated form of the electron configuration.

The use of symbol of the previous noble gas as part of the electron configuration of an element is the shortcut way to write the electron configuration.

Xenon has electron configuration as 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6}.

Cesium has the electron configuration as 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 6s^{1}

So, Cesium can be abbreviated using the symbol of xenon [Xe] 6s^{1}.

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Svetach [21]

Answer:H2=11.4g

CH4=28.6g

Explanation:The complete combustion of the two gases can be represented by a balanced reaction below

1. CH4 +2O2___CO2+2H2O

2.2H2+O2___2H2O

Combining the two we have CH4 +2H2+3O2___

CO2+4H2O

Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.

Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g

Mass of CH4=20/28×40=28.6g

2H2=8/28×40=11.4g

7 0
3 years ago
Why does the same side of the moon always face earth? A. The rotational period of earth is the same as that of the moon B. The r
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What is density? How do you find it?
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2 years ago
The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding
slega [8]

Answer:

Binding\ energy=43.43\times 10^{-20}\ J

Explanation:

Using the expression for the photoelectric effect as:

E=h\nu_0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

\nu_0=\frac {c}{\lambda_0}

Applying the equation as:

\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Given, \lambda=4.00\times 10^{-7}\ m

\frac {hc}{\lambda_0} is the binding energy or threshold energy

\frac {1}{2}\times m\times v^2 is the kinetic energy of the electron emitted.  = 6.26\times 10^{-20}\ J

Thus, applying values as:

\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy

\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

Binding\ energy=43.43\times 10^{-20}\ J

5 0
3 years ago
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