Why are 1-chlorobutane and 2-chlorobutane structural isomers?.

Give the name and formula of the compound formed from the following elements: (a) sodium and nitrogen; (b) oxygen and strontium;

lisov135 [29]

Answer:

For a: The chemical name and chemical formula formed is sodium nitride and $Na_3N$

For b: The chemical name and chemical formula formed is strontium oxide and $SrO$

For c: The chemical name and chemical formula formed is aluminium chloride and $AlCl_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

For a: Sodium and nitrogen:

Sodium is the 11th element of periodic table having electronic configuration of $[Ne]3s^1$

To form $Na^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for sodium nitride is $Na_3N$

For b: Oxygen and strontium

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium oxide is $SrO$

For c: Aluminium and chlorine

Aluminium is the 13th element of periodic table having electronic configuration of $[Ne]3s^23p^1$

To form $Al^{3+}$ ion, this element will loose 3 electrons.

Chlorine is the 17th element of the periodic table having electronic configuration of $[Ne]3s^23p^5$

To form $Cl^{-}$ ion, this element will gain 1 electron.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for aluminium chloride is $AlCl_3$

4 0

4 years ago

The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic a

hichkok12 [17]

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.

8 0

3 years ago

The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v).

vovangra [49]

Answer:

Density is an intensive property

Explanation:

The intensive properties don´t depend on mass or size, if you have a liter of water, this litter have a density equal 1000 kg/m^3. If zo have 250 liters of water, the density is 1000 kg/m^3 too.

The density does not change with the addition or substraction of matter.

I hope I have been clear.

8 0

3 years ago

Read 2 more answers

Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL

Scorpion4ik [409]

Basis: 1 L of the substance.
(1.202 g/mL) x (1000 mL) = 1202 g
mass solute = (1202 g) x 0.2 = 240.2 g
mass solvent = 1202 g x 0.8 = 961.6 g
moles KI = (240.2 g) x (1 mole / 166 g) = 1.45 moles
moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent
= 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
= 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
= 1.45 moles / (1.45 moles + 53.42 moles) = 0.0264

5 0

4 years ago

Consider the following intermediate chemical equations.

Degger [83]

Answer:

-1,300 kJ

I don't want to explain it brainly AAAAA

4 0

3 years ago