Chemistry
27.12.2020 01:58
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Your doorbell is malfunctioning, It will only strike the bell one time when the

Your doorbell is malfunctioning, It will only strike the bell one time when the doorbell is pressed. How does the electromagnet play a role in the correct (or incorrect) function of your doorbell?
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ash9951
ash9951
4,6(11 marks)

Electromagnet plays a very important role by creating a magnetic field which will in turn pull the armature towards it, thereby causing the hammer to strike the bell.

Explanation:

An electromagnet is simply a type of magnet where the magnetic field is produced with the help of an electric current.

In an electric bell, an electromagnet forms a very vital part alongside the armature, armature rod, spring, hammer and a gong.

Now, power that produces electric current in the electric bell is gotten when the switch for the bell is turned on.

However, hen the power that produces the electric current for the electromagnet stops, the magnetic field will stop being produced but when electric current is flowing through the coils, the electromagnet will create a magnetic field which will in turn pull the armature towards it, thereby causing the hammer to strike the bell.

kleathers97
kleathers97
4,7(79 marks)

Approximately 4.92.

Explanation:

Initial volume of the solution: V = 190.0\; \rm mL = 0.1900\; \rm L.

Initial quantity of \rm NH_3:

\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}.

Ammonia \rm NH_3 reacts with hydrochloric \rm HCl acid at a one-to-one ratio:

\rm NH_3 + HCl \to NH_4 Cl.

Hence, approximately n({\rm HCl}) = 0.154375\; \rm mol of \rm HCl\! molecules would be required to exactly react with the \rm NH_3\! in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that 0.3733\; \rm mol \cdot L^{-1}\rm HCl solution required for reaching the equivalence point of this titration:

\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}.

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately 0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L.

If no hydrolysis took place, 0.154375\; \rm mol of \rm NH_4 Cl would be produced. Because \rm NH_4 Cl\! is a soluble salt, the solution would contain 0.154375\; \rm mol\! of \rm {NH_4}^{+} ions. The concentration of \rm {NH_4}^{+}\! would be approximately:

\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}.

However, because \rm NH_3 \cdot H_2O is a weak base, its conjugate \rm {NH_4}^{+} would be a weak base.

\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}.

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}.

Let x\; \rm mol \cdot L^{-1} be the increase in the concentration of \rm H^{+} in this solution because of this reversible reaction. (Notice that x \ge 0.) Construct the following \text{RICE} table:

\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}.

Thus, at equilibrium:

Concentration of the weak acid: [{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M.Concentration of the conjugate of the weak acid: [{\rm NH_3}] = x\; \rm M.Concentration of \rm H^{+}: [{\rm {H}^{+}}] \approx x\; \rm M.

\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}.

\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}

Solve for x. (Notice that the value of x\! is likely to be much smaller than 0.255782. Hence, the denominator on the left-hand side (0.255782 - x) \approx 0.255782.)

x \approx 1.19929 \times 10^{-5}.

Hence, the concentration of \rm H^{+} at the equivalence point of this titration would be approximately 1.19929 \times 10^{-5}\; \rm M.

Hence, the pH at the equivalence point of this titration would be:

\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}.

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