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2 years ago

Can 1750 mL of water dissolve 4.6 moles of Copper Sulfate CuSO4? _________ Why? / Why not?

Chemistry

1 answer:

Wewaii [24]2 years ago

8 0

Answer:

No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.

Explanation:

This question is part of a Post-Lab exercise sheet.

Such sheet include the saturation concentrations for several salts.

The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.

That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.

Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.

You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.

1. Volume in liters:

V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter

2. Molar concentration, molarity, M:

M = number of moles of solute / volume of solution in liters

M = 4.6 moles / 1.75 liter = 2.6 M

Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.

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Part 1 : Answer is only B substance is soluble in water.

In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.

Part 2 : Substance A is not soluble in water and substance B is soluble in water.

According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.

But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.

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2 years ago

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2 years ago

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of

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Ideal solutions obey Raoult's law, which states that:

P_i = x_i*(P_pure)_i

where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i

In this case,

P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr

The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:

P_total = (439.6 + 118.9) torr = 558.5 torr

Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:

118.9 torr/558.5 torr = 0.213

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2 years ago

Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be

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The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

HCOOH (formic acid) → HCOO⁻ (formate) + H⁺

H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺

Formal Charges:

Formal charges are calculated using following formula,

F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

F.C = [6] - [6 + 2/2]

F.C = [6] - [6 + 1]

F.C = 6 - 7

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2 years ago

What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi

kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

$\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s)$ .

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

$\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq)$ .

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

How to rewrite a chemical equation to produce a net ionic equation?

Rewrite all reactants and products that ionizes completely in the solution as ions.
Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

Soluble salts,
Strong acids, and
Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as $\rm Mg^{2+}$ and $\rm 2\; Cl^{-}$ .
NaOH is a strong base. Each unit of NaOH can be written as $\rm Na^{+}$ and $\rm OH^{-}$ .
Mg(OH)₂ is a weak base and should not be written.
NaCl is a salt and is soluble in water. Each unit of NaCl can be written as $\rm Na^{+}$ and $\rm Cl^{-}$ .