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scoray [572]
3 years ago
6

The table above shows the number of protons in an Adam of four different elements according to the periodic table which of these

elements is a metal

Chemistry
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

<u>The correct answer is X</u>

<u />

Explanation:

First have some idea about : Valency , Metals and non- metals.

Valency : It is the combining capacity of elements . The elements combine to other elements to gain stability . They always try to complete their octet(8 electrons in the outermost shell)

For Example : Na has 11 electron

Its electronic configuration is : 2 ,8 ,1 (2+8+1 = 11)

First shell contain maximum = 2 electron

2nd shell = maximum 8 electron

3 shell = maximum 18 electron

When it remove 1 electron from outermost shell , it is left with 2 ,8 configuration. Now , its outer most shell has 8 electron . so it get stable and show +1 valency.

When the element looses electron , it get positive charge

When the element gain electron it get negative charge

Metals: those element which show +1 , +2 , +3 valency are considered as metals

Non- metals = Those metals which show the valency of -1 , -2 ,-3 are considered as Non- Metals.

Questions:

1. T = 7 proton

Electronic configuration = 2 , 5

It gains (-ve charge) 3 electron to achieve complete octet(5+3 = 8) = -3 valency

It is non- metal

2. Q=15

Its electronic configuration = 2 ,8 , 5

It gains (-ve charge) 3 electron to achieve complete octet(5+3 = 8) = -3 valency

It is non- metal

3.X = 12

Its electronic configuration = 2 ,8 , 2

It loses (+ve charge) 2 electron to achieve complete octet( 2 , 8) = + 2 valency

It is  metal

4.Z = 17

Its electronic configuration = 2 ,8 , 7

It gains (-ve charge) 1 electron to achieve complete octet(7+1 = 8) = -1 valency

It is non- metal

<u>There is only one metal = X</u>

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In covalent bonds, atoms share their electrons to reach 8 valence electrons.  Example: In water (H2O), Oxygen shares one valence electron with one atom of hydrogen, and another valence electron with another atom of hydrogen.  Oxygen now has 8 (4 unshared + 2 of its own + 1 from hydrogen + 1 from hydrogen), and each hydrogen has 2 valence electrons: one of its own and one from oxygen [ note that hydrogen only needs 2 valence electrons to be complete instead of 8].

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3 years ago
A scientist measures the standard enthalpy change for the following reaction to be 81.1 kJ :2CO2(g) + 5 H2(g)C2H2(g) + 4 H2O(g)B
posledela

<u>Answer:</u> The enthalpy of the formation of CO_2(g) is coming out to be -410.8 kJ/mol.Z

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

For the given chemical reaction:

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The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ

Putting values in above equation, we get:

81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol

Hence, the enthalpy of the formation of CO_2(g) is coming out to be -410.8 kJ/mol.

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