Answer is: b. more than 7.
The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.
For example, basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).
Balanced chemical reaction of acetic acid and sodium hydroxide:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).
Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.
The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
Learn more about heat transfer:
brainly.com/question/6363778
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Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is
