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jolli1 [7]
1 year ago
7

for small molecules and ions, arrange the intermolecular forces according to their relative strengths.

Chemistry
1 answer:
Wewaii [24]1 year ago
3 0

According to small molecules and ions, we order the intermolecular forces according to their relative strengths:

  1. Ion-Ion forces
  2. Ion-Dipole forces
  3. Hydrogen bonding
  4. Dipole-Dipole forces
  5. London's dispersion forces

This order is correct because of the nature of the different forces.

<h3>What is the nature of the various forces mentioned above?</h3>

  • Ion-ion forces are the strongest because they involve the electrostatic attraction between two ions that have opposite charges.
  • Ion-dipole forces involve the attraction between a charged ion and a polar molecule, which is weaker than the ion-ion forces.
  • Hydrogen bonding is a particularly strong form of dipole-dipole interaction that is stronger than regular dipole-dipole forces.
  • Lastly, London dispersion forces are the weakest of the forces because they are caused by temporary dipoles that form due to the random motion of electrons within a molecule.

For small molecules and ions, arrange the intermolecular forces according to their relative strengths.

List them from stronger to weaker:

- Ion-ion forces

- Hydrogen bonds

- Dipole-dipole forces

- Ion-dipole forces

- London dispersion forces

Learn more about intermolecular forces:

brainly.com/question/2193457

#SPJ4

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konstantin123 [22]

The factors that affect the rate of solution formation are capable of either increasing the rate of solution formation or decreasing it.

<h3>Factors that affect the rate of solution formation</h3>

The rate of solution formation from a chemical reaction depends on the following factors:

  • Quantity of solute and solvent: The solute should be in less quantity and solvent should be more in quantity to dissolve the solute. This will enhance the formation of solution

  • The size of the solute particles: When the solute particles are in smaller sizes, they react faster with the solvent to form solutions.

  • The temperature of solvent: In increased temperature, the solute reacts faster with the solvent.

Therefore, the factors that affect the rate of solution formation are capable of either increasing the rate of solution formation or decreasing it.

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Cold climates favor mechanical weathering.

Explanation:

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3 years ago
You are given a stock solution of 500.0 mL of 1.00M magnesium chloride solution. Calculate the volume of the stock solution you
alexgriva [62]

Answer:

50\; \rm mL of the stock solution would be required.

Explanation:

Assume that a solution of volume V contains a solute with a concentration of c. The quantity n of that solute in this solution would be:

n = c \cdot V.

For the solution that needs to be prepared, c = 0.20\; \rm M = 0.20\; \rm mol \cdot L^{-1}. The volume of this solution is V = 250.0\; \rm mL. Calculate the quantity of the solute (magnesium chloride) in the required solution:

\begin{aligned}n &= c \cdot V \\ &= 0.20\; \rm mol \cdot L^{-1} \times 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^{-1} \times 0.2500\; \rm L \\ &= 0.050\; \rm mol\end{aligned}.

Rearrange the equation n = c \cdot V to find an expression of volume V, given the concentration c and quantity n of the solute:

\displaystyle V= \frac{n}{c}.

Concentration of the solute in the stock solution: c(\text{stock}) = 1.00\; \rm M = 1.00\; \rm mol \cdot L^{-1}.

Quantity of the solute required: n = 0.050\; \rm mol.

Calculate the volume of the stock solution that would contain the required n = 0.050\; \rm mol of the magnesium chloride solute:

\begin{aligned}& V(\text{stock}) \\ &= \frac{n}{c(\text{stock})} \\ &= \frac{0.050\; \rm mol}{1.00\; \rm mol \cdot L^{-1}} \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}.

3 0
3 years ago
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