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Harrizon [31]
2 years ago
9

Electrons are always shared evenly during covalent bonding. O True False

Chemistry
1 answer:
neonofarm [45]2 years ago
8 0

Answer:

False

Explanation:

There are two types of covalent bonding: polar covalent boding and nonpolar covalent bonding. During polar covalent bonding, the electrons are not shared evenly. This is due to the electronegativity differences between the two atoms. During nonpolar covalent bonding, the electrons are shared evenly.

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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and al
Musya8 [376]

Answer:

The value of K_p is 0.02495.

Explanation:

Initial concentration of SCL_2 gas = 0.675 M

Initial concentration of C_2H_4 gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)

initially

0.675 M            0.973 M        0

At equilibrium ;

(0.675-0.35) M            (0.973-2 × 0.35) M        0.35 M

The equilibrium constant is given as :

K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}

=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}

K_c=14.45

The relation between K_p and K_c are :

K_p=K_c\times (RT)^{\Delta n}

where,

K_p = equilibrium constant at constant pressure = ?

K_c = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

\Delta n = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}

K_p=6.2\times 10^{4}

K_p=0.02495

The value of K_p is 0.02495.

7 0
3 years ago
An automobile is driving uphill. Which form of energy is not involved in this process?
Lapatulllka [165]
The answer is D) Electromagnetic energy. This is not a type of energy that occurs when a vehicle is moving.
7 0
4 years ago
Waters states of matter include steam liquid water and ice. What about water is the same in the states? What can you conclude ab
julia-pushkina [17]
This lesson is the first in a three-part series that addresses a concept that is central to the understanding of the water cycle—that water is able to take many forms but is still water. This series of lessons is designed to prepare students to understand that most substances may exist as solids, liquids, or gases depending on the temperature, pressure, and nature of that substance. This knowledge is critical to understanding that water in our world is constantly cycling as a solid, liquid, or gas.

In these lessons, students will observe, measure, and describe water as it changes state. It is important to note that students at this level "...should become familiar with the freezing of water and melting of ice (with no change in weight), the disappearance of wetness into the air, and the appearance of water on cold surfaces. Evaporation and condensation will mean nothing different from disappearance and appearance, perhaps for several years, until students begin to understand that the evaporated water is still present in the form of invisibly small molecules." (Benchmarks for Science Literacy<span>, </span>pp. 66-67.)

In this lesson, students explore how water can change from a solid to a liquid and then back again.

<span>In </span>Water 2: Disappearing Water, students will focus on the concept that water can go back and forth from one form to another and the amount of water will remain the same.

Water 3: Melting and Freezing<span> allows students to investigate what happens to the amount of different substances as they change from a solid to a liquid or a liquid to a solid.</span>
5 0
3 years ago
Which group or groups of atoms are the only atoms with f orbitals A. All atoms heavier than barium B. All atoms heavier than kry
Mashcka [7]

Answer:

All atoms heavier than barium

Explanation:

In the periodic table, elements are divided into blocks. We have the;

s- block elements

p- block elements

d- block elements

f- block elements

However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.

Hence elements heavier than barium all possess f-orbitals.

6 0
4 years ago
Fe + H2SO4 = Fe2(SO4)3 + H2
a_sh-v [17]
C (667) that’s the answer boiii
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3 years ago
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