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Sonja [21]
1 year ago
9

Table B: Enthalpy of Formation of Reactants and Products

Chemistry
1 answer:
Inessa [10]1 year ago
5 0

The calculated enthalpy values are as follows:

  • Total enthalpy of reactants = -103.85 KJ/mol
  • Total enthalpy of products = -2057.68 KJ/mol
  • Enthalpy of reaction = -1953.83 kJ/mol

<h3>What is the enthalpy of the reaction?</h3>

The enthalpy of the reaction is determined as follows:

  • Enthalpy of reaction = Total enthalpy of products -Total enthalpy of reactants
  • Total enthalpy of reactants = (ΔHf of Reactant 1 x Coefficient) + (ΔHf of Reactant 2 x Coefficient)
  • Total enthalpy of products= (ΔHf of Product 1 x Coefficient) + (ΔHf of Product 2 x Coefficient)

Equation of reaction equation: C₃H₈ (g) + 5 O(g) → 4 H₂O(g) + 3CO₂(g)

Total enthalpy of reactants = (-103.85 * 1) + (0 * 5)

Total enthalpy of reactants = -103.85 + 0

Total enthalpy of reactants = -103.85 KJ/mol

Total enthalpy of products = (-393.51 * 4) +(-241.82 * 3)

Total enthalpy of products = (-1574.04) + (-483.64)

Total enthalpy of products = -2057.68 KJ/mol

Enthalpy of reaction =  -2057.68 KJ/mol -(-103.85 KJ/mol)

Enthalpy of reaction = -1953.83 kJ/mol

In conclusion, the enthalpy of the reaction is determined from the difference between the total enthalpy of products and reactants.

Learn more about enthalpy of reaction at: brainly.com/question/14047927

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Answer:

a. slows diffusion

Explanation:

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Balancing Al+ Agcl, i have problems with this so help!!!
EleoNora [17]

Answer:

Al  + 3AgCl  →  AlCl₃  + 3Ag

Explanation:

The given equation is:  

      Al  + AgCl  →  

We are to find the product and hence balance the equation. This problem is a simple single replacement reaction.

By virtue of this, Aluminum will displace Ag from the solution:

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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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