The ΔG°f of atomic oxygen is 230.1 kJ/mol. Find ΔG° for the following dissociation reactionO2 (g) <--> 2O (g)then calculate its equilibrium constant at 298 K.

Answers

Answer 1
Answer:

Answer:

Kc = 2.145 × 10⁻⁸¹

Explanation:

Let's consider the following reaction:

O₂(g) ⇄ 2O(g)

The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:

ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)

where,

ni are the moles of products and reactants

ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants

In this case,

ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)

ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol

ΔG° = 460.2 kJ/mol

With this information, we can calculate the equilibrium constant (Kc) using the following expression:

Kc=e^(-\Delta G \°/R.T ) = e^{-460.2 kJ/mol/(8.314 * 10^(-3)kJ/mol.K)  * 298K }=2.145 * 10^(-81)


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How many grams of Cl are in 31.2g CF2Cl2

Answers

Answer:

Mass = 42.6 g

Explanation:

Given data:

Mass of CF₂Cl₂ = 31.2 g

Mass of Cl₂ = ?

Solution:

Number of moles of CF₂Cl₂ = mass/molar mass

Number of moles =  31.2 g/121 gmol

Number of moles = 0.3  mol

1 mole of CF₂Cl₂ contain 2 moles of Cl atom.

0.3 mol × 2 = 0.6 mol

Mass of Cl:

Mass = number of moles × molar mass

Mass = 0.6 mol × 71 g/mol

Mass = 42.6 g

Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4

Answers

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C, -161.5 °C  and -87.7 °C, clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

Draw a well-labelled diagram showing how your body digests food​

Answers

i too used it

i thought it will help

nice time. .....

Explanation:

...................

How is the combined gas law modified to form the ideal gas law?A) each side of the equations is divided by the number of moles
B) the temperature of the gas is stipulated to be absolute zero
C) each side of the equation is multiplied by air pressure at sea level
D) the volume of each gas particle is added to the equation

Answers

The answer is A) each side of the equations is divided by the number of moles
The combined gas law equation
     P1V1/T1 = P2V2/T2
becomes
     P1V1/n1T1 = P2V2/n2T2
with the addition of Avogadro's law relating volume and number of moles of a gas.This modification is when the number of moles of gas in a sample is allowed to change in addition to pressure, temperature and volume. This also means that pressure multiplied by the volume and divided by the product of the number of moles and temperature is a constant:
     PV/nT = constant
which if written in the form PV = nRT is the ideal gas law equation where R is the gas constant.

Answer:

A) each side of the equations is divided by the number of moles

Explanation:

If the patient has to be administered a dosage of 2 tablets every 8 hours for 7 days,what is the number of tablets required for the prescribed dosage

Answers

42 tablets. 2*8*7=42

Which equation represents the reaction of a weak acid with water

Answers

Answer:

Which equation represents the reaction of a weak acid with water?  the equation is : HCl + H2O H3O+ + Cl- HCO3– + H2O H2CO3 + OH– H2O H + + OH- HCOOH + H2O H3O+ + HCOO

Explanation:

hope it helps : )

Answer:

Hey mate, here is your answer. Hope it helps you.

HCOOH + H2O ↔ H3O+ + HCOO-

Explanation:-

A strong acid is one which completely dissociates into its corresponding ions in aqueous medium.

In contrast,  a weak will only partially dissociate such that there is an equilibrium between the dissociated ions and the undissociated acid.

In the given examples:

HCl, HCO3- and H2CO3 are all strong acids. However, HCOOH i.e. formic acid is a weak acid which dissociates in water to form H3O+ and formate ion, HCOO-

HCOOH + H2O ↔ H3O+ + HCOO-