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2 years ago

Describe how a covalent bond can involve a sharing of the electrons, but still develop charges on the atoms

Chemistry

1 answer:

adelina 88 [10]2 years ago

6 0

Answer:

Explanation:

Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms to gain more stability, which is gained by forming a full electron shell. By sharing their outermost (valence) electrons, atoms can fill up their outer electron shell and gain stability.

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Question # 10

satela [25.4K]

The answer should be D, cells can only create an identical copy of the original cell.

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3 years ago

How many electrons are in the n = 2 shell of a magnesium atom?

Montano1993 [528]

Answer:

eight

Explanation:

That means there are 12 electrons in a magnesium atom

hope this helps

8 0

2 years ago

Which of the following is a property of carbon atoms?

marusya05 [52]

Answer:

B. They can for covalent bonds with other atoms.

Explanation:

Carbon = King of the elements on the periodic table

For its proclivity to form stable covalent bonds with multivalent atoms.

5 0

3 years ago

Read 2 more answers

Determine the root-mean square speed of CO2 molecules that have an average kinetic energy of 4.2 x10-21 J per molecule.

strojnjashka [21]

<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2

If you substitute into this formula, you will get out the root-mean-square speed.

If energy is Joules, the mass should be in kg, and the speed will be in m/s.

1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg.

If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule.

4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg

So, if E = 1/2 mv^2

</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68
Take the square root of that, and you get the answer 339 m/s.

8 0

3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago