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cricket20 [7]
3 years ago
13

What is it that lives if it is fed, and dies if you give it a drink

Chemistry
2 answers:
nadezda [96]3 years ago
6 0

Answer:

ants if you splash a ant with sum water that shít is dead

Svetach [21]3 years ago
4 0

Answer:

fire

Explanation:

You might be interested in
If you have 5.42 x 1024 aluminum atoms, approximately how many moles is that
sattari [20]

Answer:

<h2>9.00 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.42 \times  {10}^{24} }{6.02 \times {10}^{23} }  \\  = 9.0033

We have the final answer as

<h3>9.00 moles</h3>

Hope this helps you

4 0
3 years ago
What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.

A) Reaction with NaI :

Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .

The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)

NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.

1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane

The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)

B) Reaction with AgNO3 :

Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.

AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )

The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.

7 0
3 years ago
What controls the amount of product formed in a chemical reaction
melisa1 [442]

Answer: limiting reactant controls the amount of product formed in a chemical reaction.

* Hopefully this answers your question :) Mark me the brainliest:)

~ 234483279c20~

5 0
3 years ago
The mass of 2.50 moles of calcium fluoride is ____ grams
Vesna [10]

Answer:

195.187016

Explanation:

3 0
3 years ago
Read 2 more answers
Which tool would give the most precise measurement of 15 milliliters of water
vagabundo [1.1K]

A graduated cylinder

_________________

o0o0o0o0o0o0o0o0o0

6 0
2 years ago
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