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3 years ago

Which process would require the least amount of energy?

1 answer:

8 0

Answer: a) the process for the first ionization energy

Every time you take a electron you're requiring more and more energy. Expelling the first one will require less energy than expelling the second and the second will require less than the third, and so on.
When you take the first one, the atom becomes positive and with that the negative forces of the electron will be more attracted to the positive charge. The more electrons that are lost, the more positive this ion will become, causing it to be more difficult to separate the electrons from the atom.

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galvanic cell is powered by the following redox reaction: (g) (aq) (l)(aq) (s) (aq) Answer the following questions about this ce

kotykmax [81]

Answer:

Explanation:

1. the 1/2 reaction that occurs at the cathode

3Cl2(g) +6e^- -------------> 6Cl^- (aq)

2 the 1/2 reaction that occurs at the anode

2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-

E0 = -0.59v

3Cl2(g) +6e^- -------------> 6Cl^- (aq)

E0 = 1.39v

3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)

E0cell = 0.80v

5 0

3 years ago

Which of the following is an example of a

Sergio039 [100]

Answer:

dissolving sugar into water is a physical change

5 0

2 years ago

A cough sytup contains 0.5M dextromethophan. How many moles of the cough supressant are in 21.3mL of the cough syrup?

Doss [256]

Answer:

0.0107 mol

Explanation:

Multiply concentration by volume (in liters) to get moles.

0.5 M • 0.0213 L = 0.0107 mol

7 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte in

Vika [28.1K]

Explanation:

Apple cider vinegar or balsamic vinegar might contain various synthetics which might interact with or nullified the findings, whereas white vinegar comprises of acidic acid only. In fact, apple cider vinegar and balsamic vinegar are deep in appearance this would make it very difficult to determine the color. This is why it is preferable to use white vinegar in the Titration process.

3 0

3 years ago