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The table shows a chemical equation and the dimensional analysis used by a student to calculate the number of moles of Ba3N2 required to produce 6.7 moles of Ba(OH)2. Student's Dimensional Analysis Chemical Equation Dimensional Analysis Ba3N2 + 6H2O → 3Ba(OH)2 + 2NH3 3 mol of Ba(OH)2 x 1 mol of Ba3N2 / 6.7 mol of Ba(OH)2 What is the error in the dimensional analysis?

Answers

Answer 1

Answer:

Answer:

Its arrangement.

Explanation:

Hello,

In this case, we rewrite the chemical reaction as:

$Ba_3N_2+6H_2O-->3Ba(OH)_2+2NH_3$

Now, a more organized dimensional analysis is shown below to determine the very same required value:

$n_(Ba_3N_2)=6.7molBa(OH)_2*(1molBa_3N_2)/(3molBa(OH)_2)$

Without giving the numerical answer, one sees that the student's dimensional analysis is wrongly written as long as the 3 moles of $Ba(OH)_2$ must be dividing rather than multiplying as it is the relationship factor used to relate it with the $Ba_3N_2$ based on the reaction stoichiometry. Moreover, the 6.7moles of $Ba(OH)_2$ must be multiplying as it is the starting value.

Best regards.

Answer 2

Answer:

Answer:

The error is in the rule of three.

3 mol of Ba(OH)2 ___ comes from ____ 1 mol Ba3N2

6.7 mol of Ba(OH)2 ___ comes from ___ (6.7 .1) / 3

Explanation:

Ba3N2 + 6H2O → 3Ba(OH)2 + 2NH3

3 mol of Ba(OH)2 x 1 mol of Ba3N2 / 6.7 mol of Ba(OH)2

- The rule of three is wrong.

- The chemical equation is ballanced.

- The question is not clear about which is the limiting reactant. May be we don't have enough water, so that's the limiting and Ba3N2 is in excess. So have to work with H2O.

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Answers

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Answers

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Answer:

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Explanation:

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Answers

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Answers

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Explanation:

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^(0) - [((0.0592)/(n) · log Q)]$

where

E: Cell potential (non standard conditions)

$E^(0)$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = (C^(c) * D^(d) )/(A^(a)*B^(b) )$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^(0)$ , n and Q

Let´s calcule potential in nomal conditions ( $E^(0)$ ):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions: Eo (v):

$(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ -0.23

$Ni --> Ni^(2+)$ + 2 e- +0.99

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions: Eo (v):

2 * [ $(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ ] -0.23

1 * [ $Ni --> Ni^(2+)$ + 2 e- ] +0.99

------------------------------------------------------------------- -------------

2 $(VO_(2))^(2+)$ + 2e- + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$ + 2e- 0.76 v

Then global balanced chemical reaction is:

2 $(VO_(2))^(2+)$ + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$

and the potential in nomal conditions is:

$E^(0)$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = ([(VO_(2)+]^(2)* [Ni^(2+) )/([VO_(2+)]^(2) * Ni )]$

From the exercise we know:

$[VO_(2) ^(2+)] = 2.5 M$

$[VO_(2)+] = 0.083 M$

$[Ni^(2+)] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^(0) - ((0.0592)/(n) - log Q)\n \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v$

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

Learn more about Nernst Equation here:

brainly.com/question/31593791

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