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4 years ago

Choose all the answers that apply MORE THAN ONE ANSWER. Salt is an ionic compound. It _____.

Chemistry

2 answers:

Anuta_ua [19.1K]4 years ago

7 0

dissolves easily in water

has a high boiling point

has a high melting point

conducts electricity when melted

Explanation:

Salts are ionic compounds. Ionic compounds are formed when two atoms exchange their valence electrons. One atom gains and the other atom loses.

Generally, ionic compounds have the following properties:

They are usually hard solids with high melting points or liquids with high boiling points.
They are soluble in water and insoluble in non-polar solvents. This is conforms with the solubility rule that like dissolve like.
In aqueous solution or molten form, ionic compounds conducts electricity.
They take part in very fast chemical reactions.

Learn more:

Ionic compounds brainly.com/question/6071838

#learnwithBrainly

iris [78.8K]4 years ago

6 0

dissolves easily in water

has a high boiling point

has a high melting point

conducts electricity when melted

Hope this helps:]

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1 year ago

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dimulka [17.4K]

The answer would be 4.

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3 years ago

How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?

algol [13]

Answer:

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = $\frac{13.0 g}{162.5 g/mol}=0.08 mol$

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

$3\times 0.08 mol=0.24 mol$ of chloride

$Na^++Cl^-\rightarrow NaCl$

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

$\frac{1}{1}\times 0.24 mol=0.24 mol$ of NaCl

1 mole = $N_A=6.022\times 10^{23}$ molecules/ atoms

Number of NaCl molecules in 0.24 moles :

$=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules$

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

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3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$