Time Remaining: 1:27:31 If 50.0 g of H₂ and 100.0 g of O₂ react, how many moles of H₂O can be produced in the reaction below? 2

Response:water

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Based on the equation:

ΔG = ΔH - TΔS = 0

It follows that ΔS = ΔH/T

So, ΔS = n*ΔHVap / Tvap

- where n represents the number of moles calculated as mass/molar mass

For a mass of 24.1 g

and a molar mass of 187.3764 g/mol

substituting gives:

∴ n = 24.1 / 187.3764g/mol

      = 0.129 moles

The molar enthalpy of vaporization, ΔHvap, is 27.49 kJ/mol

The temperature in Kelvin, Tvap = 47.6 + 273 = 320.6 K

After substitution, we compute ΔS, the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

      = 11 J/K

Answer:

B, D

Explanation:

We need to recognize that the ice will rise in temperature from -6.5 ºC to 0 ºC for it to change into water.

Let's define q₁ as the heat needed to warm the ice to 0ºC, and q₂ as the heat for the transition from solid to liquid.

The calculation for q₁ is as follows:

q₁ = s x m x ΔT, where s represents the specific heat of ice (2.09 J/gºC), m is the mass, and ΔT is the temperature difference.

For q₂, the enthalpy of fusion is computed as:

q₂ = C x ΔT

with C indicating the specific heat for the phase transition, denoted as AH in kJ/mol.

All necessary data for computing q₁, q₂, and the total heat change (q₁ + q₂) is provided.

q₁ = 25.0 g x (2.09 J/gºC) x (0 - (-6.5 ºC))

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

Combining these values gives us qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ.

Now we can answer the question:

(a) False, AH refers to the heat capacity during melting.

(b) True, as we concluded earlier.

(c) False, there’s only one phase transition from solid (ice) to liquid.

(d) True based on our calculations above.

(e) False, according to our findings.

There is an attachment providing the solution.

A triprotic acid is a type of Arrhenius acid that has the ability to donate three protons per molecule during dissociation in aqueous solutions. Thus, the chemical reaction, as outlined in the question, at the third equivalence point, can be expressed as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R denotes the counter ion of the triprotic acid. Consequently, the ratio of reacted acid to base at this point is 1:3.
The moles of NaOH are calculated as 0.106M*0.0352L = 0.003731 mole. Therefore, the amount of H3R is 0.003731mole/3=0.001244mole.
Subsequently, the molar mass of the acid can be determined: 0.307g/0.001244mole=247 g/mol.