What is the molality of the solution?! :O

Chemistry

1 answer:

alisha [4.7K]3 years ago

5 0

Answer:

-200

Explanation:

My sister told me to put dat but dont put dat as your answer

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Fundamental substances that cannot be broken down chemically into simpler subsances are______

Gemiola [76]

Answer:

A. Elements

Elements are substances that cannot be separated into simpler substances. Salt is made up of the elements sodium and chloride. Water is made up of the elements hydrogen and oxygen, to list a few.

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3 years ago

An aluminum ion has a charge of +3, and an oxide has a charge of -2. What would be the product of a reaction between these two e

kozerog [31]

The product of a reaction between these two elements is $Al_{2} O_{3}$ .

Explanation:

The oxidation state of an ion in a compound is equal to its charge.

The aluminum having a charge of +3 because oxidation state is +3

The oxide is having charge of -2

The product of these reactants will produce a chemical compound.

The compound formed is $Al_{2} O_{3}$ i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.

The +3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. 

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3 years ago

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$