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3 years ago

When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.

Chemistry

1 answer:

Lubov Fominskaja [6]3 years ago

3 0

Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

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?A scientist is investigating how the lunar cycle affects the number of sea turtles nesting each night. Which of the following i

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Number of nesting turtles

Explanation:

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3 years ago

The percent of remaining parent isotope in a radioactive decay process is 40 percent. How many half-lives have elapsed since the

muminat

Answer: Between 1 and 2.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$a=\frac{a_o}{2^n}$ ............(1)

where,

a = amount of reactant left after n-half lives = 40

$a_o$ = Initial amount of the reactant = 100

n = number of half lives

Putting in the values we get:

$40=\frac{100}{2^n}$

$2^n=2.5$

taking log on both sides

$nlog(2)=log(2.5)$

$n=1.32$

Thus half-lives that have elapsed is between 1 and 2

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3 years ago

T’Keyah puts salt in ice water and then in boiling water to see which will dissolve faster.Which dissolving rate factor is she t

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Answer:

temperature

Explanation:

its in the book

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3 years ago

Read 2 more answers

what is the maximum amount of moles of P2O5 that can theoretically be made from 136 g of P4 and excess oxygen

zhuklara [117]

We are given with
136 g P4
excess oxygen

The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles

Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5

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3 years ago

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Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest

balu736 [363]

Answer:

Cost to supply enough vanillin is $3.2\$$

Explanation:

Threshold limit of vanillin in air is $2.0\times 10^{-11}g$ per litre means there should be $2.0\times 10^{-11}g$ of vanillin in 1L of air to detect aroma of vanillin.

$1ft^{3}=28.32L$

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So amount of vanillin should be present to detect = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$

As cost of 50 g vanillin is $112\$$ therefore cost of $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$ vanillin = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$$

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3 years ago