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avanturin [10]
2 years ago
15

Which of the following occurs when a covalent bond forms? Which of the following occurs when a covalent bond forms? Partial char

ges on polar molecules interact. Electrons in valence shells are shared between atoms. Nonpolar molecules are pushed together by surrounding water molecules. Electrons in valence shells are transferred from one atom to another.
Chemistry
1 answer:
Alexus [3.1K]2 years ago
7 0

Answer: Option (b) is the correct answer.

Explanation:

A covalent bond is defined as the bond which occurs due to sharing of electrons between the combining atoms.

Generally, a covalent bond is formed between non-metals.

For example, both nitrogen and oxygen atoms are non-metals and they combine covalently to form NO_{2} compound.

As nitrogen has 5 valence electrons and an oxygen atom has 6 valence electrons. So, there occurs unequal sharing of electrons between the two.

Thus, we can conclude that when a covalent bond forms then electrons in valence shells are shared between atoms.

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The electron dot diagram for a neutral atom of chlorine (atomic number 17) is shown below.
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Answer:

A. 35Cl1-

Explanation:

Chlorine needs 1 more electron to have full octet thus will take 1 electron and possess a -1 charge.

7 0
2 years ago
Katie needs to measure water for a recipe she is using to cook dinner she should measure the liquid in...
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Answer:

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3 years ago
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2 years ago
A car with a mass of 1,200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. The centri
Helga [31]
The answer is 80 m.

Centripetal force (F) is a force that makes body move around a circular curve. The unit of force is N (N = kg * m/s²).
It can be represented as:
F= \frac{m* v^{2} }{r}
where:
m - mass
v - velocity
r - radius of the curve

We have:
m = 1,200 kg
V = 20 m/s
F = 6,000 N = 6,000 kg * m/s²

We need radius of the curve:
r = ?

So, if <span>F= \frac{m* v^{2} }{r}, then:
</span>    r= \frac{3* v^{2} }{F}
⇒ r =  \frac{1200 kg * (20m/s)^{2} }{6000 kg*m/s^{2} }
⇒ r= \frac{1200*400*kg*m^{2}*s^{2}  }{6000kg*m/s^{2} }
⇒ r= \frac{480,000m}{6000}
⇒ r=80m


6 0
3 years ago
How many oxygen molecules are in 22.4 liters of oxygen gas at 273k and 101.3kpa?
babunello [35]

How many oxygen molecules are in 22.4 liters of oxygen gas at 273k and 101.3kpa

First solve the number of moles of the oxygen gas by using the ideal gas equation:

PV = nRT

Where n is the number of moles

n = PV/RT

n = (101 300 Pa) (22.4 L) (1 m3/1000 L ) / ( 8.314 Pa m3 / mol K) ( 273 K)

n = 1 mol O2

the number of molecules can be solve using avogrados number 6.022x10^23 molecule / mole

molecules of one mole O2 = 6.022x 10^23 molecules

8 0
2 years ago
Read 2 more answers
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