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3 years ago

The maximum amount of solute that dissolves in a given amount of solvent and forms a stable solution is called the

Chemistry

1 answer:

denis23 [38]3 years ago

3 0

Answer:

The maximum amount of solute that dissolves in a given amount of solvent and forms a stable solution is called the solubility of the solute

Explanation:

The maximum amount of solute that could be dissolved in a given amount of solvent is the solubility of the solute. It is the saturated solution's concentration from where a saturated solution can be defined as the one which already contains the maximum quantity of dissolved solute at a specified temperature, while an unsaturated solution is one with a capacity to dissolve more solutes

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Given a balanced chemical equation below: 3Cu(s) + 2H3PO4 --- > Cu3(PO4)2 + 3H2 How many moles of copper are needed to react

Harman [31]

Answer:

7.5 moles

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2

From the balanced equation above,

3 moles of Cu reacted with 2 moles of H3PO4.

Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e

Xmol of Cu = (3 x 5)/2

Xmol of Cu = 7.5 moles

Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.

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3 years ago

What elements are found in the compound represented in the diagram

DedPeter [7]

hydrogen and carbon, hope that helped

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4 years ago

What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?

rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

5 0

3 years ago

After a big game, you put your water bottle inside the refrigerator. The next

12345 [234]

Mostly and for what I would say is A

6 0

3 years ago

Calculate the molarity of sodium ion in a solution made

Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

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

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

<u> The final concentration of sodium anion = 0.1035 M</u>

4 0

3 years ago