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2 years ago

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When an ionic compound dissolves in water: When an ionic compound dissolves in water: the solvent-solute attractive forces overc

ome the solute-solute attractions. the positive end of water dipoles attract the negative ions. the negative end of water dipoles attract the positive ions. each of the above (A, B, and C) occurs. none of the above (A, B, or C) occurs.

1 answer:

oksano4ka [1.4K]2 years ago

8 0

Answer:

each of the above (A, B, and C) occurs

Explanation:

When an ionic compound dissolves in the water, the following happens :

-- the solvent solute attractive forces tries to overcome the solute solute attractions.

-- the water dipoles' negative end attracts the positive ions

-- the water dipoles' positive end attracts the negative ions

For example,

NaCl which is an ionic compound and also a strong electrolyte, it dissociates into water on the hydrated Na cations as well as Cl anions.

In water, the oxygen has negative charge and thus attracts the positive ions of the sodium, whereas the hydrogen is of positive and it attract the ions of chlorine which is negative.

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Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (

Nesterboy [21]

Balanced chemical equation for the reaction is:

2S $O_{2}$ (g) + $O_{2}$ (g)+ 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

Moles of $H_{2} S_{} O_{4}$ formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of S $O_{2}$ = 11.5 moles

moles of $H_{2} S_{} O_{4}$ = ?

Moles of $O_{2}$ needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2S $O_{2}$ (g) + $O_{2}$ (g) + 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of S $O_{2}$ , x moles of $H_{2} S_{} O_{4}$ is formed

$\frac{1}{2} =\frac{x}{11.5}$

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

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2 years ago

A 50 kg object _____.

stich3 [128]

Answer:

will have more mass the moon than the earth

Explanation:

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2 years ago

Li2SO4 _____ an electrolyte in solution.<br> A. Is <br> B. Is not

Rasek [7]

Answer:

a

Explanation:

it is an electrolyte because of its strong polar chemical bond

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2 years ago

Read 2 more answers

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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3 years ago

4Al(s) + 30₂(g) → 2Al2O3 (s)

Margaret [11]

1) 1 molecules
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