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2 years ago

12

If you were a given a beaker containing a clear solution (with no solid solute at the bottom), what simple test would you do to

determine if it is saturated or unsaturated solution?

1 answer:

raketka [301]2 years ago

6 0

Answer:

put a filter to see if there is any stagering solution

Explanation:

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0

2 years ago

8.43g of CaCl2·2H2O is dissolved in 40.0g of H2O. What is the molality of the solution? (Note that the H2O in the CaCl2·2H2O mus

lara31 [8.8K]

Answer:

[CaCl₂·2H₂O] = 1.43 m

Explanation:

Molality is mol of solute / kg of solvent.

Mass of solvent = 40 g

Let's convert g to kg → 40 g / 1000 = 0.04 kg

Let's determine the moles of solute (mass / molar mass)

8.43 g / 146.98 g/mol = 0.057 mol

Molality = 0.057 mol / 0.04 kg → 1.43

3 0

3 years ago

Read 2 more answers

NaHCO 3 and NaHSO 4 are acid salts. however aqueous solution of NaHCO 3 is basic while aqueous solution of NaHSO 4 is acidic. gi

posledela

Answer:c

Explanation:

3 0

3 years ago

What can be said about an exothermic reaction with a negative entropy change?.

pashok25 [27]

Spontaneous at low temperatures.

3 0

2 years ago

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

Maurinko [17]

<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressure 1
V₁ is volume 1
P₂ is pressure 2
V₂ is volume 2

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
[Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
[Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
[Pressure] Rewrite: P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0

3 years ago