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irinina [24]
3 years ago
13

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Not all entry-fields provided may be needed. Choose "na" in any space where you have no reagent. A. H2, Ni B. H2, Pd/CaSO4, quinoline C. H2, Ni2B(P-2) D. Br2 E. NaNH2, Mineral oil, heat F. NH4Cl G. t-BuOK, t-BuOH, heat H. CH3I I. CH3CH2I
Chemistry
1 answer:
aliya0001 [1]3 years ago
3 0

Answer:

Explanation:1. NaNH2 (1-Butene)

CH3CH2CH2CH2Cl --------------> CH3CH2CH=CH2 + HCl (elimination reaction)

2. Br2, CCl4

CH3CH2CH=CH2 ---------------> CH3CH2CH(Br)CH2Br (Simple addition Reaction)

3. NaNH2 (1-Butyne)

CH3CH2CH(Br)CH2Br ----------------> CH3CH2C≡CH + 2HBr

Sodamide (NaNH2) is a very strong base and generally results in Terminal Alkynes when treated with Vicinal Dihalides.

Alcoholic KOH on the other hand results in the formation of Alkynes with triple bonds in the middle of the molecule.

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How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

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3 years ago
What is the the correct equation for water gas​
nikdorinn [45]

Answer:

Water-gas A mixture of carbon monoxide (CO) and hydrogen (H2) produced by passing steam over red-hot coke using the endothermic reaction C + H2O → CO + H2.

6 0
3 years ago
Relative dating determines the relative age of fossils according to
KiRa [710]
The answer Is C.the layer of rock they're found in
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Use molecular structure and intermolecular bonding to describe why bromine has a lower boiling point than water
kotegsom [21]

Answer:

Bromine mollecules are held together by van der waals forces while a water molecule constitutes both van der waals forces and hydrogen bomnding

Explanation:

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3 0
3 years ago
"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"
arlik [135]

Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

5 0
3 years ago
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