1) Describe the characteristics of molecular compounds that affect their particular physical

1 answer:

8 0

The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.

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How is the kinetic energy of the constituent particles of any matter affected by a change in temperature?

SpyIntel [72]

Answer:

Particles in all states of matter are in constant motion and this is very rapid at room temperature. A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. ... Individual particles in liquids and gases have no fixed positions and move chaotically.

Explanation:

5 0

3 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

Write the formulas of the three singly chlorinated isomers formed when 2,2-dimethylbutane reacts with Cl2 in the presence of lig

Tems11 [23]

The molecule with same molecular formula but different arrangement of atoms is said to be an isomer.

When 2,2-dimethylbutane reacts with chlorine in the presence of light gives three isomers that is $CH_{3}C(CH_{3})_{2}CHClCH_{3}$ (3-chloro-2,2-dimethylbutane), $ClCH_{2}C(CH_{3})_{2}CH_{2}CH_{3}$ (1-chloro-2,2-dimethylbutane) and $ClCH_{2}CH_{2}C(CH_{3})_{2}CH_{3}$ (1-chloro-3,3-dimethylbutane).

In above case, the molecular formula of all isomers are same i.e. $C_{6}H_{13}Cl$ but chlorine is arranged in different positions of carbon. Thus, results isomers.