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Virty [35]
3 years ago
12

Write the balanced molecular equation for the neutralization reaction between hcl and ba(oh)2 in aqueous solution. include physi

cal states.
Chemistry
2 answers:
mr_godi [17]3 years ago
5 0
Each Ba(OH)2 requires 2 Cl to produce one BaCl2 molecule. 

<span>The 2 H's from the 2 HCl molecules combine with the 2 (OH) to form 2 H2O molecules. </span>

<span>2 HCl (aq) + Ba(OH)2 (s) ==> BaCl2 (aq) + 2 H2O (l)

I hope the answer will help you. </span>
Elena L [17]3 years ago
4 0

<u>Answer:</u> The balanced molecular equation is written below

<u>Explanation:</u>

A molecular equation is the chemical equation in which the ionic compounds are written as molecules rather than component ions.

Neutralization reaction is defined as the chemical reaction when an acid reacts with a base to produce a salt and water molecule.

The balanced molecular equation for the reaction of hydrochloric acid and barium hydroxide follows:

2HCl(aq.)+Ba(OH)_2(aq.)\rightarrow BaCl_2(aq.)+2H_2O(l)

By Stoichiometry of the reaction:

2 moles of aqueous solution of hydrochloric acid reacts with 1 mole of aqueous solution of barium hydroxide to produce 1 mole of aqueous solution of barium chloride and 2 moles of liquid water.

Hence, the balanced molecular equation is written above.

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Distinguish between deliquescence and efflorescence. ​
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Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
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(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

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  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

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