Answer:
No, it is not enthalpy favored since the chemical system gains energy.
Explanation:
The dissolution of ammonium nitrate in water is an endothermic process.
Endothermic process requires the system to gain energy to can dissolve the particles in water.
So, the reaction is not enthalpy favored.
The pH of the monoprotic weak acid is 2.79.
<h3>What are weak acids?</h3>
The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.
The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c (monoprotic acid) = 0.33 M.
Ka = 1.2·10⁻⁶
[A⁻] = [H⁺] = x
[HA] = 0.33 M - x
Ka = [A⁻]·[H⁺] / [HA]
2. 6 × 10⁻⁶ = x² / (0.33 M - x)
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺]
pH = -log(0.000524 M)
pH = 2.79
Thus, the pH of the monoprotic weak acid is 2.79
To learn more about weak acids, refer to the below link:
brainly.com/question/13032224
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Answer:
The answer to your question is P = 1.357 atm
Explanation:
Data
Volume = 22.4 L
1 mol
temperature = 100°C
a = 0.211 L² atm
b = 0.0171 L/mol
R = 0.082 atmL/mol°K
Convert temperature to °K
Temperature = 100 + 273
= 373°K
Formula
![(P + \frac{a}{v^{2}} )(v - b) = RT](https://tex.z-dn.net/?f=%28P%20%2B%20%5Cfrac%7Ba%7D%7Bv%5E%7B2%7D%7D%20%29%28v%20-%20b%29%20%3D%20RT)
Substitution
![(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)](https://tex.z-dn.net/?f=%28P%20%2B%20%5Cfrac%7B0.211%7D%7B22.4%7D%29%2822.4%20-%200.0171%29%20%3D%20%280.082%29%28373%29)
Simplify
(P + 0.0094)(22.3829) = 30.586
Solve for P
P + 0.0094 = ![\frac{30.586}{22.3829}](https://tex.z-dn.net/?f=%5Cfrac%7B30.586%7D%7B22.3829%7D)
P + 0.0094 = 1.366
P = 1.336 - 0.0094
P = 1.357 atm
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
I think it is C. coefficients but I don't know what subscripts are