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earnstyle [38]
3 years ago
6

Which compound is an exception to the octet rule? H20 HCI CCI4 CIF3

Chemistry
1 answer:
Reika [66]3 years ago
8 0

Answer:

clf3

Explanation:

it occupied more than 8 valence electrons

You might be interested in
"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"
Artyom0805 [142]

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

4 0
3 years ago
What effect does pneumonia have on the diffusion of O2 and CO2 across the membranes in the lungs? a. slows diffusion b. complete
GarryVolchara [31]

Answer:

a. slows diffusion

Explanation:

Gas exchange on respiratory surfaces in the body (the lungs) occurs through a process known as diffusion. Blood which is low in oxygen and high in carbondioxide (carried from cells) goes through an exchange in the lung's alveoli (where oxygen concentration is high and carbondioxide is low). The oxygen in the alveoli diffuses into the blood, while the carbondioxide in the blood diffuses into the alveoli. This diffusion is possible because of the concentration gradient across the membranes.

Pneumonia is the inflammation of the lungs due to injury or infection. Liquid (pus) accumulates in the alveoli (a natural immune response to the infection or injury), a condition known as pulmonary edema which makes it harder for gases to be exchanged between the blood and the alveoli, thereby making breathing difficult. This slows down diffusion and if the condition is severe enough, can cause a respiratory failure where oxygen levels in the blood are critically low and carbondioxide levels are very high.

5 0
3 years ago
Here is the electron configuration for Magnesium. How many total electrons are in the 2nd energy level?
balandron [24]

Answer:

8 electrons

Explanation:

Magnesium is present on group 2.

It has 2 valence electrons.

Electronic configuration of magnesium:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

1st energy level contain 2 electrons.(1s²)

2nd energy  level contain 8 electrons. (2s² 2p⁶)

3rd energy level contain 2 electrons. (3s²)

3rs energy level of magnesium is called valence shell. It contain two valance electrons. Magnesium can easily donate its two valance electrons and get stable electronic configuration.

It react with halogens and form salt. For example,

Mg + Cl₂   →   MgCl₂

7 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
3. Mr. Hill has 27 students in his class
stich3 [128]
Just dived both numbers by two
4 0
3 years ago
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