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laiz [17]
4 years ago
6

When a weak base is titrated with a strong acid, the ph at the equivalence point is?

Chemistry
1 answer:
Rainbow [258]4 years ago
6 0
NA depiction of the pH change during a titration of HCl solution into an ammonia solution. The curve depicts the change in pH (on the y-axis) vs. the volume of HCl added in mL (on the x-axis). In strong acid-weak base titrations, the pH at the equivalence point is not 7 but below it.
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Which of the following is an example of the conversion of electrical energy to electromagnetic energy?
GuDViN [60]
The ccorrect answer is C a generator tubrine
6 0
4 years ago
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Chlorine (CI) can be found as an ion with a 1- charge. How does this happen?
Lesechka [4]

Answer:There are 18 electrons and 17 protons, so the chlorine atom has become a charged chlorine ion with a charge of negative one (-1). ... When it does, the sodium atom becomes a sodium ion with a charge of positive one (+1). Chlorine, as mentioned above, desperately wants an electron so it can fill its outer electron level.

8 0
3 years ago
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The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
quester [9]

Answer:

Ksp=1.07x10^{-8}

Explanation:

Hello,

In this case, the dissociation reaction is:

PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)

For which the equilibrium expression is:

Ksp=[Pb^{2+}][I^-]^2

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[Pb^{2+}]=1.39x10^{-3}M

[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M

Thereby, the solubility product results:

Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}

Regards.

6 0
4 years ago
Can someone help me? It needs to have a diagram that has arrows.
daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
Measure the initial temperature of the water to the nearest 0.1°C. Record in the data table.
Aleksandr [31]

Answer:

b

Explanation:

8 0
3 years ago
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