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NikAS [45]
3 years ago
8

Match the following family of elements to its group.

Chemistry
1 answer:
docker41 [41]3 years ago
7 0

Explanation:

To solve this problem, we simply use the periodic table of elements which groups elements based on their atomic numbers.

The atomic number of an element is the number of protons it contains. The protons are the positively charged particles within an atom.

  • The vertical arrangement of elements on the periodic table is the group.
  • The horizontal arrangement of elements is the period.

Now;

Noble gases  belongs to group 18

Alkali earth metals belongs to group 2

Halogens belongs to group  17

Alkali metals belongs to group 1

Transition metals belongs to group 3-12

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According to VSEPR theory, if there are three electron domains in the valence shell of an atom, they will be arranged in a(n) __
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Answer:  B. trigonal planar

Explanation:

Valence Shell Electron Pair Repulsion Theory (VSEPR) is a model to predict the geometry of the atoms making up a molecule where atoms are arranged such that the forces of repulsion are minimum.

If a central atom is bound to three electron domains ,the number of electron pairs is 3, that means the hybridization will be sp^2 and the electronic geometry of the molecule will be trigonal planar as the electron pairs will repel each other and attain a position which is most stable.

Example: BCl_3 with 3 electron domains has trigonal planar geometry.

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Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<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.

The equation used to calculate enthalpy change is of a reaction is:

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

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

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

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

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