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levacccp [35]
4 years ago
11

What kind of chemical change do you think occurs when a banana peel turns brown in the open air?

Chemistry
1 answer:
Alisiya [41]4 years ago
6 0
Rotting, and ageing. The banana peel ages faster in open air, and therefor begins to rot. Brow peels don't mean that the banana is rotted, however, it is a sign that it may begin to rot soon.
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The Answer you're looking for here is Star A mainly because it's the star with the highest Surface Temperature on the Graph.

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The swallowtail caterpillar is an example of a ?
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B)  Consumer COnsumer COnsumer

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3 years ago
a 25 ml sample of gas is in a syringe at 22C if it was cool down to 0 C what will the volume of the gas become
Tpy6a [65]

Answer: 0.023 liters

Explanation:

Given that,

Original volume of gas (V1) = 25mL

[convert 25mL to liters

If 1000ml = 1L

25ml = 25/1000 = 0.025L]

Original temperature of gas (T1) = 22°C

[Convert 22°C to Kelvin by adding 273

22°C + 273 = 295K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 0°C

[Convert 0°C to Kelvin by adding 273

0°C + 273 = 273K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

0.025L/295K = V2/273K

To get the value of V2, cross multiply

0.025L x 273K = 295K x V2

6.825L•K = 295K•V2

Divide both sides by 295K

6.825L•K/295K = 295K•V2/295K

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4 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.

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3 years ago
Why are planets spherical?
damaskus [11]

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Gravity pulls all sides equally forming a ball

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