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DiKsa [7]
2 years ago
11

How do you find the # of sigma and pi bonds using a Lewis dot structure

Chemistry
1 answer:
Lubov Fominskaja [6]2 years ago
3 0

Every single bond is a sigma bond.

A double bond contains 1 sigma bond and 1 pi bond.

A triple bond contains 1 sigma bond and 2 pi bonds.


For example, in acetylene, H-C≡C-H, the sigma bond is formed by the head-on overlap of two sp orbitals. The pi bonds are formed by the side-on overlap of 2p orbitals.

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Zn + I2 ---&gt; Znl2<br> Determine the theoretical yield of the product if 2g of Zn is used
egoroff_w [7]

Answer:

Mass = 9.58 g

Explanation:

Given data:

Mass of Zn = 2g

Theoretical yield of ZnI₂ = ?

Solution:

Chemical equation:

Zn + I₂       →     ZnI₂

Number of moles of Zn:

Number of moles = mass/molar mass

Number of moles = 2g / 65.38 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of Zn and ZnI₂.

                   Zn           :          ZnI₂

                    1             :           1

                  0.03        :         0.03

Mass of ZnI₂:

Mass = number of moles × molar mass

Mass = 0.03 mol × 319.22 g/mol

Mass = 9.58 g

4 0
2 years ago
We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
2 years ago
The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
QveST [7]

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

3 0
3 years ago
Which of these groups CANNOT harvest energy from the sun?
Tomtit [17]
Viruses..................
8 0
3 years ago
Read 2 more answers
What type of plate boundary is<br> shown in the diagram below?
ryzh [129]
Answer : The second one
8 0
2 years ago
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