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weeeeeb [17]
4 years ago
6

Why does the solution decolorize on standing after the equivalence point has been reached?

Chemistry
2 answers:
Arisa [49]4 years ago
8 0

The equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant.

Depending on the concentration of titrant we could be adding little excess of it and this may result in persistence of color of solution. After continuous stirring for a while the excess titrant may react with dissolved CO₂ in air and thus decolorizing the solution.

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Colt1911 [192]4 years ago
3 0

Answer:

The potassium permanganate slowly fades in color as it is dissolved in the water.

Explanation:

Hello,

Perhaps, your talking about a titration that was carried out by using potassium permanganate as the titrant, thus, bringing to bear the fact that the equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant, one could state that as the titration is carried out, the potassium permanganate slowly fades in color as it is dissolved in the water after the equivalence point is reached since the formed species become colorless as well.

Best regards.

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A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

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Bill sets up an experiment to determine and predict the velocity of a marble at the end of a track. He uses an electronic balanc
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Answer:

A

Explanation:

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2 years ago
1) A light bulb takes in 30 of energy per second. It transfers 3j as use
natta225 [31]

Answer:

\boxed{\text{10 \%}}

Explanation:

The formula for efficiency is  

\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}

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Useful energy =  3 J

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Calculation:

\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}

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Hmmmmmm, Gamma Radiation 
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Just wanted to make sure I have the right idea.
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Explanation:

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