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nikitadnepr [17]
3 years ago
5

Of the molecules below, only ________ is polar.a. SiCl4 b. CH4 c. CCl4 d. SeF4

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
6 0

Answer:

SeF4 is a polar molecule

Explanation:

SeF4 is a polar molecule because a polar molecule is any molecule that have lone pairs of electrons in the central atom or have atoms that are electronegative and the electrons between that are covalently bonded are not evenly distributed.

The electronegative atoms of flourine in SeF4 are not evenly distributed and kind pairs of electrons are on the central atom.

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For each of the following sets of elements, circle the element expected to be most electronegative and draw a box which is expec
Evgen [1.6K]

Answer: The element expected to be most electronegative is Ca.

The element expected to be least electronegative is K.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

Down the group:

The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

As, the size of an element increases, the valence electrons gets away from the nucleus. So, the attraction between the nucleus and the shared pair of electrons decreases

Hence, electronegativity decreases moving from top to bottom down a group

Across a period:

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases.

Hence, electronegativity increases moving across left to right in a period.

Thus as K, Sc and Ca are arranged across a period, the electronegativity order is K< Sc < Ca.

7 0
3 years ago
A base is a substance that accepts protons. true or false?
umka21 [38]
True................
3 0
2 years ago
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There are two different compounds of sulfur and fluorine.
olga nikolaevna [1]
What's your question......................
5 0
3 years ago
For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does,
jolli1 [7]

Answer:

Lowers the actual yield

Explanation:

5 0
3 years ago
What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
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