(e) what oxidation number is assigned to nitrogen in the reactant nh₄cl? (f) is this reaction a redox reaction? (g) what is the molarity of the nh₄cl before it is mixed with the agno₃? (h) imagine that the nh₄cl needed to be diluted to make 50 ml of 0.3 m nh₄cl. what volume, in milliliters, of the original nh₄cl is needed for this dilution? (i) what mass of agno₃ would be required to react completely with 100 ml of 0.20m nh₄cl? (j) ki is dissolved in water to make it aqueous. then, o₂ gas is bubbled through the solution. assume a reaction takes place. write the balanced equation and predict the products for the reaction. (k) what atom is oxidized? assign oxidation numbers as evidence to show your work. (l) the product containing potassium is undergoes a secondary reaction. it reacts with the water. will it form an acid or base upon reaction? (m) the reaction described above is accidentally contaminated with h₂s. what is the name of this contaminant? (n) the ph after the contamination is found to be 4.2. what is the h⁺? (o) the concentration of the ki is unknown, but its absorbance is measured in a spectrophotometer at 0.32. (assume that a is 13,000 m⁻¹cm⁻¹ and the depth of the cuvette was 1 cm. what is the concentration of ki?)

Question

Accepted Answer

### (e)
# Explanation:
## Step1: Recall oxidation - number rules for nitrogen in $NH_4Cl$
In $NH_4Cl$, the ammonium ion $NH_4^+$ has a charge of + 1. Chloride ion $Cl^-$ has a charge of - 1. In the ammonium ion, hydrogen has an oxidation number of + 1. Let the oxidation number of nitrogen be $x$.
## Step2: Set up the equation for the sum of oxidation numbers in $NH_4^+$
For the $NH_4^+$ ion, we have $x+(+1)\times4 = + 1$.
## Step3: Solve the equation for $x$
$$
\begin{align*}
x + 4&=1\
x&=1 - 4\
x&=-3
\end{align*}
$$
# Answer:
The oxidation number of nitrogen in $NH_4Cl$ is - 3.

### (f)
# Explanation:
## Step1: Write the reaction between $NH_4Cl$ and $AgNO_3$
The reaction is $NH_4Cl+AgNO_3
ightarrow AgCl\downarrow + NH_4NO_3$. The mole - ratio of $NH_4Cl$ to $AgNO_3$ is 1:1.
## Step2: Calculate the moles of $AgNO_3$
The volume of $AgNO_3$ is $V = 100\ mL=0.1\ L$ and its molarity $M = 0.20\ M$. The moles of $AgNO_3$, $n(AgNO_3)=M\times V=0.20\ mol/L\times0.1\ L = 0.02\ mol$.
## Step3: Calculate the moles of $NH_4Cl$ needed
Since the mole - ratio of $NH_4Cl$ to $AgNO_3$ is 1:1, the moles of $NH_4Cl$ needed, $n(NH_4Cl)=0.02\ mol$.
## Step4: Calculate the molarity of the diluted $NH_4Cl$ solution
The final volume of the $NH_4Cl$ solution after dilution is $V_{final}=50\ mL = 0.05\ L$. Using the formula $M=\frac{n}{V}$, the molarity of the diluted $NH_4Cl$ solution, $M=\frac{0.02\ mol}{0.05\ L}=0.4\ M$.
## Step5: Use the dilution formula $M_1V_1 = M_2V_2$ to find the initial volume of $NH_4Cl$
We know $M_1 = 0.3\ M$, $M_2 = 0.4\ M$, and $V_2 = 50\ mL$. Rearranging the formula $V_1=\frac{M_2V_2}{M_1}$.
$$
\begin{align*}
V_1&=\frac{0.4\ M\times50\ mL}{0.3\ M}\
V_1&=\frac{20}{0.3}\ mL\
V_1&\approx66.7\ mL
\end{align*}
$$
# Answer:
The volume of the original $0.3\ M\ NH_4Cl$ solution needed is approximately $66.7\ mL$.

### (i)
# Explanation:
## Step1: Write the reaction between $NH_4Cl$ and $AgNO_3$
The reaction is $NH_4Cl + AgNO_3
ightarrow AgCl\downarrow+NH_4NO_3$. The mole - ratio of $NH_4Cl$ to $AgNO_3$ is 1:1.
## Step2: Calculate the moles of $AgNO_3$
The volume of $AgNO_3$ is $V = 100\ mL = 0.1\ L$ and its molarity $M = 0.20\ M$. The moles of $AgNO_3$, $n(AgNO_3)=M\times V=0.20\ mol/L\times0.1\ L=0.02\ mol$.
## Step3: Calculate the moles of $NH_4Cl$ needed
Since the mole - ratio of $NH_4Cl$ to $AgNO_3$ is 1:1, the moles of $NH_4Cl$ needed, $n(NH_4Cl)=0.02\ mol$.
## Step4: Calculate the moles of $NH_4Cl$ in the original solution
The volume of the original $NH_4Cl$ solution is $V_1$ (to be found), and its molarity $M_1 = 0.3\ M$. The moles of $NH_4Cl$ in the original solution $n_1 = M_1V_1$. After dilution to $V_2 = 50\ mL=0.05\ L$ with a molarity $M_2$ (calculated in part (f) as $0.4\ M$), the moles of $NH_4Cl$ remain the same.
## Step5: Calculate the mass of $AgNO_3$
The molar mass of $AgNO_3$ is $M(AgNO_3)=107.87\ g/mol + 14.01\ g/mol+3\times16.00\ g/mol=169.88\ g/mol$. The moles of $AgNO_3$ is $n = 0.02\ mol$. Using the formula $m=n\times M$, the mass of $AgNO_3$ is $m = 0.02\ mol\times169.88\ g/mol = 3.3976\ g\approx3.40\ g$.
# Answer:
The mass of $AgNO_3$ required to react completely with $100\ mL$ of $0.20\ M\ NH_4Cl$ is approximately $3.40\ g$.

### (j)
# Explanation:
## Step1: Write the reaction between $KI$ and $O_2$ in aqueous solution
The reaction is $4KI + O_2+2H_2O
ightarrow 2I_2 + 4KOH$.
## Step2: Predict the products
The products of the reaction are iodine ($I_2$) and potassium hydroxide ($KOH$). The oxygen in $O_2$ is reduced and the iodine in $KI$ is oxidized.
# Answer:
The balanced equation is $4KI + O_2+2H_2O
ightarrow 2I_2 + 4KOH$, and the products are $I_2$ and $KOH$.

### (k)
# Explanation:
## Step1: Recall the rules for assigning oxidation numbers
In $KI$, potassium has an oxidation number of + 1. Let the oxidation number of iodine be $x$. For a neutral compound $KI$, $+1 + x=0$, so $x=-1$. In $I_2$, the oxidation number of iodine is 0.
## Step2: Determine the oxidized atom
The oxidation number of iodine changes from - 1 in $KI$ to 0 in $I_2$. Since the oxidation number of iodine increases, iodine is the atom that is oxidized.
# Answer:
The atom that is oxidized is iodine.

### (l)
# Explanation:
## Step1: Write the reaction between potassium and water
Potassium reacts with water according to the equation $2K + 2H_2O
ightarrow 2KOH+H_2\uparrow$. In this reaction, potassium is oxidized and water is reduced. The base formed is potassium hydroxide ($KOH$).
# Answer:
The base formed is $KOH$.

### (m)
# Explanation:
## Step1: Identify the reaction between $H_2S$ and the contaminant
The reaction between $H_2S$ and the contaminant (assuming it is an oxidizing agent) can lead to the formation of sulfur. The contaminant that reacts with $H_2S$ is likely an oxidizing agent. The name of the contaminant could be an oxidizing agent. If we consider a metal - containing contaminant, it could be a metal ion in a high oxidation state. For example, if the contaminant is a metal ion like $Fe^{3 +}$, the reaction could be $2Fe^{3+}+H_2S
ightarrow 2Fe^{2 +}+S\downarrow + 2H^+$. The contaminant $H_2S$ reacts with is an oxidizing agent.
# Answer:
The name of the contaminant is an oxidizing agent.

### (n)
# Explanation:
## Step1: Recall the relationship between pH and $[H^+]$
The pH is defined as $pH =-\log[H^+]$. Given $pH = 4.2$.
## Step2: Solve for $[H^+]$
$$
\begin{align*}
4.2&=-\log[H^+]\
\log[H^+]&=- 4.2\
[H^+]&=10^{-4.2}\ M\
[H^+]&\approx6.31\times10^{-5}\ M
\end{align*}
$$
# Answer:
The $[H^+]$ is approximately $6.31\times10^{-5}\ M$.

### (o)
# Explanation:
## Step1: Recall the Beer - Lambert law
The Beer - Lambert law is $A=\epsilon cl$, where $A$ is the absorbance, $\epsilon$ is the molar absorptivity, $c$ is the concentration, and $l$ is the path - length. Given $A = 0.32$, $\epsilon=13000\ M^{-1}cm^{-1}$, and $l = 1\ cm$.
## Step2: Solve for the concentration $c$
$$
\begin{align*}
0.32&=13000\ M^{-1}cm^{-1}\times c\times1\ cm\
c&=\frac{0.32}{13000\ M^{-1}}\
c&\approx2.46\times10^{-5}\ M
\end{align*}
$$
# Answer:
The concentration of $KI$ is approximately $2.46\times10^{-5}\ M$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

(e)

Step1: Recall oxidation - number rules for nitrogen in \(NH_4Cl\)

Step2: Set up the equation for the sum of oxidation numbers in \(NH_4^+\)

Step3: Solve the equation for \(x\)

Step1: Write the reaction between \(NH_4Cl\) and \(AgNO_3\)

Step2: Calculate the moles of \(AgNO_3\)

Step3: Calculate the moles of \(NH_4Cl\) needed

Step4: Calculate the molarity of the diluted \(NH_4Cl\) solution

Step5: Use the dilution formula \(M_1V_1 = M_2V_2\) to find the initial volume of \(NH_4Cl\)

Step1: Write the reaction between \(NH_4Cl\) and \(AgNO_3\)

Step2: Calculate the moles of \(AgNO_3\)

Step3: Calculate the moles of \(NH_4Cl\) needed

Step4: Calculate the moles of \(NH_4Cl\) in the original solution

Step5: Calculate the mass of \(AgNO_3\)

Answer:

(f)