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3 years ago

Which diagram represents an element that is likely to form covalent bonds?

Physics

2 answers:

Serhud [2]3 years ago

7 0

Answer:

The second option

Explanation:

The octet rule states that elements tend to bond in such a way that each atom has eight electrons in its valence (outermost) shell, except for the first level which is complete with 2 electrons.

The first picture shows an atom with only 1 electron in its valence shell, the second picture 5 electrons, the third pictures 8 electrons and the fourth picture 2 electrons. The first and last cases are more likely to lose their valence electrons to form bonds than share them into a covalent bond (because they are few). The third case has the octet rule already satisfied, so is not likely to neither lose/receive nor share its electrons. The atom of the second case can share 3 electrons to complete the octet rule.

ankoles [38]3 years ago

3 0

Answer:

Ok look Im going try my best, and the answer that most likely looks like it can be correct is C the third picture i hope this is correct and helps bye bye:3

also do you wanna get some points >w>

Explanation:

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

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4 years ago

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

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Alona [7]

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kicyunya [14]

Answer:

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