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PilotLPTM [1.2K]
2 years ago
11

create a flowchart that outlines when to use dimensional analysis and when to use scientific notation.

Chemistry
1 answer:
murzikaleks [220]2 years ago
7 0

Answer:

  see attached

Explanation:

Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.

Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.

__

Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.

On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.

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The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

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2 years ago
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<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

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Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

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E^o_{cell}=E^o_{cathode}-E^o_{anode}

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Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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