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katrin [286]
3 years ago
5

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.

Chemistry
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

The correct option is: A. 0.168 M

Explanation:

Chemical reaction involved:

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O

Given: <u>For MnO₄⁻ solution</u>-

Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;

<u>For Fe²⁺ solution</u>:

Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M

<u><em>To find out the concentration of Fe²⁺ solution (M₂), we use the equation:</em></u>

\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}

\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}

0.505 = M_{2}\times 3

M_{2} = \frac{0.505}{3} = 0.168M

<u>Therefore, the concentration or molarity of Fe²⁺ solution: </u><u>M₂ = 0.168 M</u>

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4 0
3 years ago
List a possible set of four quantum numbers (n,l,ml,ms) in order, for the highest energy electron in gallium?
Elena-2011 [213]
Gallum: Z = 31

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Highest energy electron: 4p1

Quantum numbers:

n = 4, because it is the shell number
l = 1, it corresponds to type p orbital 
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3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
It didn't have science but the subject is science. A climate with short cool summers and long, bitterly cold winters is the
kvasek [131]
Hey there,

Question : <span>A climate with short cool summers and long, bitterly cold winters is the?

Answer : D, Tundra Climate

Hope this helps :))

<em>~Top♥</em>
</span>
6 0
3 years ago
Read 2 more answers
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