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iogann1982 [59]
3 years ago
12

What is a chemical property of matter? reactivity volume color boiling point

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

reactivity

Explanation:

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Question
Otrada [13]

Answer:

Add more pieces of zinc (2n) to the mixture

Heat the hydrochloric acid (HC) solution.

Explanation:

The reaction rate is simply the speed of the reaction. It is usually monitored by the rate at which the concentration of either of the reactants or products are changing per unit of time.

The factors affecting the rate of reactions are:

  • Nature of the reactants
  • Concentration or pressure of gases
  • Temperature of the reaction
  • Availability of catalysts
  • Sunlight

In this problem, if the concentration of the zinc increases, the rate of reaction will increase.

If the temperature of HCl increases, so also the rate of the reaction.

8 0
3 years ago
Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene
kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}, 2-bromo-<em>2</em>-methylbutane;
  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}, 2-bromo-<em>1</em>-methylbutane.

It is expected that \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3} would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions \text{H}^{+} and negatively-charged bromide ions \text{Br}^{-}.

\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbo<em>cation</em> with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3};
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}.

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has <em>three</em> carbon-carbon single bonds after the addition of the proton \text{H}^{+} as opposed to <em>two</em> as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.  

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.  

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-<em>2</em>-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.

3 0
3 years ago
A diffusion coefficient KG was determined for the diffusion of a gas A through a stagnant film B [KG = 0.88 mol//h-ft2-atm].
Angelina_Jolie [31]

Explanation:

Transfer of mass A into stagnant film B depends on the availability of driving force.

Whereas driving force is the pressure difference at the surface of A and the bulk.

As,       N_{A} \propto (P_{A1} - P_{A2})

           N_{A} = K_{G} \times (P_{A1} - P_{A2})

Therefore, putting the given values into the above formula as follows.

         N_{A} = K_{G} \times (P_{A1} - P_{A2})

                     = 0.88 \times (0.2 atm - 0.05 atm)

                     = 0.132 mol/h.ft^{2}

Thus, we can conclude that the flux of A from a surface into a mixture of A and B is 0.132 mol/h.ft^{2}

4 0
4 years ago
Calculate the percent composition for each element in each of the following compounds.
jek_recluse [69]
The answer is c because the calculations for the elements have to connect with the compounds
6 0
3 years ago
Curious Carl and his lab partner were handed a 2 liter sealed flask containing two gases, neon and argon. The partial of each ga
stiv31 [10]

Answer:

it's difficult but i think the answer is

3 0
3 years ago
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