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Drupady [299]
3 years ago
15

Why is crushed ice used for snow cones?

Chemistry
1 answer:
aleksandrvk [35]3 years ago
4 0
<span>Increasing the surface area helps dissolve more flavoring</span>
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Sodium (Na) A Homogeneous Mixture B Element C Compound D Heterogeneous Mixture
laiz [17]
Sodium is categorized as a B. Element.
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3 years ago
What do the colors produced by fireworks indicate?
ahrayia [7]
<span>Metal compounds are used since metals have characteristic flame colors </span>
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3 years ago
When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final react
velikii [3]

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

  • Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.  
  • Propagation; free radicals react with molecules to produce new free radicals and molecules.
  • Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}

\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet

\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}

\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}

\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}

\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet

Termination

\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}

8 0
3 years ago
If you have 3 g of each substance, which would take up the smallest volume?
VashaNatasha [74]

Answer:

Explanation:

f the mass of an object is 35 grams and it takes up 7 cm3 of space, calculate the density.

34 pages

8 0
2 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
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