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mel-nik [20]
3 years ago
6

............................hhh look at the picture

Chemistry
1 answer:
kifflom [539]3 years ago
5 0

Neon :

  • Number of proton = 10
  • Group 18
<h3>Further explanation</h3>

Given

Neon atom

Required

Number of proton

Group in periodic table

Solution

Atomic number of Neon : 10

Electron configuration : [He] 2s²2p⁶

Number of proton = atomic number = 10

Neon and Helium are in the same group which is group 18 (the noble gases)

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If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w
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Answer:

The pressure of the gas increased (if temperature remained constant).

The Boyle's law supports this observation.

Explanation:

The initial measurements of the gas are given as;

volume = 100 L

Pressure = 300 kpa

The second measurement is given as;

Volume = 75 L

The second reading implies that the volume of the gas has decreased. If the temperature of the gas remained constant, then the pressure must have increased according to the Boyle's law;

At constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.

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A red car and a blue car collide. What would be an example of Locard's exchange principle in this situation?
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Answer:

The blue car left paint on the red car

Explanation:

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What do you understand by valency electron and valency shell?​
jasenka [17]

Answer:

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7 0
3 years ago
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
We know that for a given reaction when the temperature increases from 100 k to 200 k the rate constant doubles. What is the acti
Alenkasestr [34]

Answer:

The activation energy of the reaction is 1.152 kJ/mol.

Explanation:

Activation energy is the minimum amount which is absorbed by the reactant molecules to undergo chemical reaction.

Initial temperature of reaction = T_1=100 K

Final temperature of reaction = T_2=200 K

Initial rate of the reaction at 100 k = K_1=k

Final rate of the reaction at 200 k = K_2=2k

Activation energy is calculated from the formula:

\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}(\frac{T_2-T_1}{T_1T_2})

R = Universal gas constant = 8.314 J/ K mol

\log\frac{2k}{k}=\frac{E_a}{2.303\times 8.314 J/K mol}(\frac{200 K-100K}{200 K\times 100K})

E_a=1,152.772 J/mol=1.152 kJ/mol

5 0
3 years ago
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