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3 years ago

Which statement is incorrect?

Chemistry

1 answer:

Katena32 [7]3 years ago

4 0

Answer:

e. When the indicator changes color and remains, 2 more drops of the solution in the buret should be added.

Explanation:

<em>Which statement is incorrect? </em>

<em>Select one: </em>

<em> a. The volume of solution in the buret should be recorded when the endpoint is reached. </em> CORRECT. At the endpoint, the titrant is expected to have completely reacted with the titrand.

<em>b. It is good practice to place the flask over a white background to better observe the color change of the solution.</em> CORRECT. Using a white background makes it easier to detect the changes in the color of the indicator.

<em>c. The endpoint is when the indicator changes color.</em> CORRECT. This is the function of an indicator.

<em>d. Phenolphthalein will go from a colorless solution to a pink solution for the addition of NaOH to an acid solution.</em> CORRECT. Phenolphthalein is colorless when pH < 8.3 and it is pink when pH > 10.0.

<em>e. When the indicator changes color and remains, 2 more drops of the solution in the buret should be added.</em> INCORRECT. When the endpoint is reached, no more titrant should be added.

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3 years ago

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium co

Anarel [89]

Answer:

Kp = 1.41 x 10⁻⁶

Explanation:

We have the chemical equation:

2 A(g) + 3 B(g)⇌ C(g)

In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):

dn= (sum moles products - sum moles reactants)

= (moles C - (moles A + moles B))

= (1 - (2+3))

= 1 - 5

= -4

We have also the following data:

Kc = 63.2

T= 81∘C + 273 = 354 K

R = 0.082 L.atm/K.mol (it is a constant)

Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:

$Kc = (RT)^{dn}$ = (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶

3 0

2 years ago

Please help! Thanks :D

Digiron [165]

Species shown in bold are precipitates.

Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃

Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃

Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃

Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃

A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.

Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.

Compare the first and last row:

Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.

As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.

Compare the second and third row:

Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.

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3 years ago

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lbvjy [14]

Answer:

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abruzzese [7]

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