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Fittoniya [83]
1 year ago
8

Give the systematic names for the formulas or give the formulas for the names:

Chemistry
1 answer:
lilavasa [31]1 year ago
4 0

Lead (II) acetate trihydrate is the systematic name for the formula Pb(C₂H₃O₂)₂ . 3H₂O.

<h3>What is Molecular Formula ?</h3>

The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.

<h3>What is Oxidation State ?</h3>

Oxidation state is also known as oxidation number. It is defined as the atom is equal to the total number of electrons which have been removed from the element in order to form chemical bond with other atom.

Now find the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O

Assume the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O be x

x + 2 × (-1) + 3 × 0 = 0

x - 2 + 0 = 0

x = 2

Oxidation state of Pb is +2 or (II)

Thus from the above conclusion we can say that The systematic name for the formula Pb(C₂H₃O₂)₂ .3H₂O is Lead (II) acetate trihydrate.

Learn more about the Molecular Formula here: brainly.com/question/15960587

#SPJ4

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AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the
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The question is incomplete, the complete question is;

AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —

aluminum bromide

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Answer:

aluminum bromide

Explanation:

Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.

We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.

Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.

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The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The he
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Answer:

ΔH =  - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of  moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from  298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH =  56.576 kJ

The equation of the given reaction can be represented as :

C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)}  \to 3CO_{2(g)} +4H_2O_{(l)}

Thus for 0.028 mol of heat liberated; ΔH =  56.576 kJ

For 1 mole of heat liberated now:

ΔH =  56.576 kJ/0.028 mol

ΔH =  2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH =  - 2020.57 kJ/mol

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