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noname [10]
3 years ago
15

The bond between iodine and fluorine in l-F would be a:

Chemistry
1 answer:
koban [17]3 years ago
4 0

Answer:

D.) polar covalent bond

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The name of an alcohol side chain is known as an alkoxy group. True or false?
mylen [45]
By itself, i don’t think so.

though, paired with a hydrogen bond, it is.

If i’m wrong, please feel free to let me know :)
6 0
2 years ago
Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole
Talja [164]

Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
6 0
3 years ago
Oxygen would acquire a stable arrangement of electrons by bonding with two atoms of which of the following elements?
crimeas [40]
<span>D.) Oxygen would acquire a stable arrangement of electrons by bonding with two atoms of "Magnesium"
[ As Mg has 2 extra electrons & their size are quite similar ]

Hope this helps!</span>
7 0
3 years ago
Read 2 more answers
The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
belka [17]

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0
3 years ago
HELP ASAP ILL GIVE BRAINLIEST!!
Dahasolnce [82]

Of course we would experience them.

3 0
2 years ago
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