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sukhopar [10]
3 years ago
10

A chemistry student is given 700. mL of a clear aqueous solution at 26.° C. He is told an unknown amount of a certain compound

X is dissolved in the solution. The student allows the solution to cool to 26.° C. The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 0.032 kg.
Required:
Using only the information above, can you calculate the solubility of X in water at 22.0°C ? If you said yes, calculate it. Be sure your answer has a unit symbol and 2 significant digits.
Chemistry
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

The correct answer is - yes, 4.57 g of solute per 100 ml of solution

Explanation:

The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.

Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.

Then solublity will be :

32. g solute / 700 ml solution = y / 100 ml solution

⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.

Thus, the answer is 4.57 g of solute per 100 ml of solution.

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The pressure of the gas = 40 atm

<h3>Further explanation</h3>

Given

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Required

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Input the value :

P₂ = P₁V₁/V₂

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A student observes that water in a pot on a burner becomes hotter explain how energy is conserved in the situation
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  • <em>Energy can be transferred from a region to another , but there will still the same overall amount of energy</em> ( energy conservation)

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8 0
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Which is stronger - the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond str
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Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0
3 years ago
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