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3 years ago

In one to two sentences, describe the process by which the ionic compound, Licl, would dissolve in the polar solvent, CH3COCH3

2 answers:

8 0

LICL as an Ionic compounds has covalent ability and they are soluble in polar solvent such as CH3COCH3 but they are insoluble in non-polar solvents. Due to their polarity, CH3COCH3 will decreases the electrostatic forces of attraction thereby resulting in free ions in aqueous solution.

Water is the known as a polar solvent that can dissolve an ionic compounds very easy and as a polar solvent, the arrangement of oxygen and hydrogen atoms in water is in bent shape.

Ionic compounds like LICL that is very polar is soluble in the polar solvent water.

Polar solvents like CH3COCH3 will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles.

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charle [14.2K]3 years ago

5 0

Answer:

Explanation:

the ability of ionic of LiCL its because of it's signification covalent character

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3 years ago

Which law was used to determine the relationship between the volume and the number of moles?

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Explanation:

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3 years ago

Cho 700ml dung dịch NaOH 0.1M vào 100ml dung dịch AlCl3 0.2M. Tính khối lượng kết tủa sau phản ứng

Licemer1 [7]

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3 years ago

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume 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 $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

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3 years ago

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Nataly_w [17]

Your answer will be A

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3 years ago