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zvonat [6]
3 years ago
9

Why is oxygen diflouride a gas at room temperature?

Chemistry
1 answer:
atroni [7]3 years ago
3 0

Answer:

Oxygen is a simple molecular structure, where individual oxygen atoms are bonded to each other by strong covalent bonds. Hence, a low amount of energy is required to overcome these weak forces and oxygen has a low boiling point. Therefore, at room temperature, oxygen is a gas. Oxygen difluoride is a colorless gas, condensable to a pale yellow liquid, with a slightly irritating odor. It is the most stable of the compounds of fluorine and oxygen, which include O,F,, O,F, and 0,F2 but nevertheless it is a strong oxidizing and fluorinating agent. Oxygen Difluoride is a colorless gas or a yellowish-brown liquid with a foul odor. Just to finally link Joseph's answer to the question, oxygen difluoride will thus change from liquid to solid state when chilled from -220°c to -230°c. The boiling point of oxygen is -182.96 degrees Celsius (under 1 standard atmosphere). This means at temperatures below that point, oxygen is a solid or a liquid, and at temperatures above that point, oxygen is a gas. So at -183 degrees Celsius, oxygen is a liquid.

Explanation:

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It would have a zero charge because it is a neutral atom. The number of electrons which is a negative charge is equal to the number of protons which is positive, they will cancel each other out hence meaning it will become neutral
4 0
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"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.

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4 0
3 years ago
Which one of the following will change the value of an equilibrium constant?
Ad libitum [116K]

Answer:

(E) changing temperature

Explanation:

Consider the following reversible balanced reaction:

aA+bB⇋cC+dD

If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:

Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)

where:

[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1

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-  <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes

- Pure solids and liquids, including solvents, are not considered for equilibrium expression.

- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc

8 0
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An endothermic reaction has
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