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zalisa [80]
2 years ago
11

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

NaOH.
Chemistry
1 answer:
Rus_ich [418]2 years ago
6 0
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH. 
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH. 
</span>
<span>After the  reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:

 </span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
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For the reaction N2(g) + 3H2(g) 2NH3(g), what will happen if hydrogen gas was removed from the reaction mixture? There will be a
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This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.

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7 0
3 years ago
Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th
stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have:  2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}

Knowing:

  • ΔH= 75.5 kJ/mol
  • H_{NO}= 90.25 kJ/mol
  • H_{Cl_{2} }= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound  the chlorine Cl₂)
  • H_{NOCl}=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - H_{NOCl}

Solving

-H_{NOCl}=75.5 kJ/mol - 2*90.25 kJ/mol

-H_{NOCl}=-105 kJ/mol

H_{NOCl}=105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0
2 years ago
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