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lisabon 2012 [21]
3 years ago
6

The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at

each other along the line between the two nuclei.
a. True
b. False
Chemistry
1 answer:
nalin [4]3 years ago
6 0

Answer:

The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>

Explanation:

The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.

<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.

Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.

Hence , the answer is true .

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How many protons does an atom of gold have?
klasskru [66]

Answer:

There are 79 protons in the nucleus of one Gold atom.

Explanation:

The number of protons of any given atom/element can be determined by the atomic number of the specific element, which is found on the periodic table.

3 0
3 years ago
Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of li
alexira [117]

Answer:

Explanation:

To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.

The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.

This can be expressed below:

        RAM = Σmₙαₙ

where mₙ is the mass of isotope n

           αₙ is the abundance of isotope n

for this problem:

RAM of Li = m₆α₆ + m₇α₇

       m₆ is mass of isotope Li-6

        α₆ is the abundance of isotope Li-6

       m₇ is mass of isotope Li-7

        α₇ is the abundance of isotope Li-7

3 0
3 years ago
Determine the mass of MgCl2 needed to create a 100. ml solution with a concentration of 3.00 M.
kiruha [24]

Explanation:

1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\  = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1}  \: g \\  = 28.5 \: g \: of \: magnesium \: chloride

5 0
3 years ago
The potential in an electrochemical cell, E, is related to the Gibb's free energy change (ΔG) for the overall cell redox reactio
Nana76 [90]

Answer:

Explanation:

As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)

ΔG = ΔG° + RTInQ

Q = 1

ΔG = ΔG°

ΔG = =nFE°

n=no of electrons transfered.

E° = 1.1v

ΔG° = -2 * 96500 * 1.10

= -212300J

ΔG° =-212.3kJ/mol

<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
5 0
3 years ago
Using only the 1h nmr spectrum of the crude nitration product, determine if the product is mostly ortho-, meta-, or para- substi
tatiyna
<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:

</span>

In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.

On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration

The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.

The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.

7 0
3 years ago
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