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quester [9]
2 years ago
10

Are the strengths of the interactions between the particles in the solute and between the particles in the solvent before the so

lute and solvent are combined greater than, less than, or equal to the strengths of the interactions between solute particles and solvent particles after dissolution? Explain.
Chemistry
1 answer:
Colt1911 [192]2 years ago
5 0

Answer:

Less than

Explanation:

The process of dissolution occurs as a kind of "tug of war". On one side are the solute-solute and solvent-solvent interaction forces, while on the other side are the solute-solvent forces.

Only when the solute-solvent forces are strong enough to overcome the pre-mixing forces do they overcome the "tug of war", and thus dissolution occurs.

Thus, it is concluded that the interaction forces between solute particles and solvent particles before they are combined are less than the interaction forces after dissolution.

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The metallic radius of a lithium atom is 152 pm. What is the volume of a lithium atom in cubic meters?
Anuta_ua [19.1K]

Answer:

Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³

Explanation:

Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,

V=\frac{4}{3}\pi  R^{3}----------------------------------------------------------------------------------------(Eq. 1)

Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters

R=(152pm )(1X10^{-12}\frac{m}{pm})

R = 1.52 X 10^{-10}m

placing this value in Eq.1 the required result is achieved

V=\frac{4}{3}\pi  {1.52X10^{-10}}^{3}

V= 1.47 X 10⁻²⁹ m³

8 0
2 years ago
Functional group and bond hybridization of vanillin
lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

7 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
The boiling point of another member of this homologous series was found to be 309 KK. What is the likely molecular formula for t
Ganezh [65]

Answer: Pentane C5H12

Explanation:

The boiling point of a substance is simply defined as the temperature whereby a liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid will changes into vapor.

The likely molecular formula for this compound is Pentane i.e C5H12 due to the fact that its boiling point is between Butane with formula C4H10 and Hexane with formula C6H14 boiling points.

8 0
2 years ago
Phosphorus has the molecular formula p4, and sulfur has the molecular formula s8. how many grams of phosphorus contain the same
Alinara [238K]
1) Find the number of molecules in 7.88 g of sulfur

molar mass of S8 = 8*atomic mass of S = 8 * 32.0 g / mol = 256.0 g/mol

Number of moles  = mass in grams / atomic mass = 7.88 g / 256.0 g / mol = 0.0308 moles

2) Find the mass of 0.0308 moles of P4

mass = number of moles * molar mass

molar mass of P4 = 4 * atomic mass of P = 4 * 31 g/mol = 124 g/mol

mass of P4 = 0.0308 moles * 124 g/mol = 3.8192g ≈ 3.82 g.

Answer: 3.82 grams of P4 will have the same number of molecules as 7.88 g of S8 (that is 0.0308 moles of molecules)
6 0
2 years ago
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