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kipiarov [429]
2 years ago
7

How much water should be added to 4.3 moles of LiBr to prepare a 2.05 m solution?

Chemistry
1 answer:
arsen [322]2 years ago
8 0

Answer:

2.1 kg of water

Explanation:

Step 1: Given data

  • Moles of lithium bromide (solute): 4.3 moles
  • Molality of the solution (m): 2.05 m (2.05 mol/kg)
  • Mass of water (solvent): ?

Step 2: Calculate the mass of water required

Molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute/kilograms of solvent

kilograms of solvent = moles of solute/m

kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg

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An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of
Artist 52 [7]

<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HBr

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL

Putting values in above equation, we get:

1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M

Hence, the molarity of Ca(OH)_2 in the solution is 0.1 M.

7 0
3 years ago
15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?
storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

FeCl₃ dissolved in 450 mL of solution(will dissociate )

Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

  • mol FeCl₃(MW=162,2 g/mol)

\tt \dfrac{15}{162.2}=0.0925

  • mol Cl⁻ :

\tt \dfrac{3}{1}\times 0.0925=0.2775

  • molarity of Cl⁻ :

\tt \dfrac{0.2775}{0.45}=0.617~M

7 0
3 years ago
Carbohydrates and proteins are built up from their basic building blocks by the ________.
user100 [1]
C, to form bonds between each monomers
4 0
3 years ago
When molecules are joined I an ordered structure​
olga2289 [7]

Answer:

ice

Explanation:

5 0
3 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Margaret [11]

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

7 0
3 years ago
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