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Mrrafil [7]
3 years ago
14

(BONUS QUESTION) In the copper complexes investigated, the water and ammonia molecules act as Lewis bases (electron pair donors)

; they provide both of the electrons that form the covalent bonds between the Cu2+ ion and each ligand. Which ligand (H2O or NH3) is the stronger Lewis base? Explain your answer in terms of your answers for questions 3a and 3b. (hint: the poorer an oxidizing agent a species is, the harder it it to reduce that species. Hint #2: the copper complexes are reduced by gaining an electron.)

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer and explanation:

Water and ammonia molecules acts as Lewis bases. its is clear from the spectrochemical series that NH₃ has stronger field ligand than water

checked thee attached image for additional solutions

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You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene. Upon complete combustion in oxygen, you collect 0.204 g o
Elan Coil [88]

Answer:

The empirical formula is C7H5N3O6  

Explanation:

Step 1: Data given

Mass of sample = 0.150 grams

Mass of CO2 = 0.204 grams

Molar mass CO2 = 44.01 g/mol

Mass of H2O = 0.030 grams

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.204 grams / 44.01 g/mol

Moles CO2 = 0.00464 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.00464 moles we have 0.00464 moles C

Step 4: Calculate mass C

Mass C = 0.00464 moles * 12.01 g/mol

Mass C = 0.0557 grams

Step 5: Calculate moles H2O

Moles H2O = 0.030 grams / 18.02 g/mol

Moles H2O = 0.00166 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00166 moles H2O we have 2* 0.00166 = 0.00332 moles H

Step 7: Calculate mass H

Mass H = 0.00332 moles * 1.01 g/mol

Mass H = 0.00335 grams

Step 8: Calculate mass N

Mass N = 0.185 * 0.150 grams

Mass N = 0.02775 grams

Step 9: Calculate moles N

Moles N = 0.02775 grams / 14.0 g/mol

Moles N = 0.00198 moles

Step 10: Calculate mass O

Mass O = 0.150 grams - 0.02775 - 0.00335 - 0.0557

Mass O = 0.0632 grams

Step 11: Calculate moles O

Moles O = 0.0632 grams / 16.0 g/mol

Moles O = 0.00395 moles

Step 11: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.00464 moles / 0.00198 moles =2.33

H: 0.00332 moles / 0.00198 moles = 1.66

N: 0.00198 moles / 0.00198 moles = 1

O: 0.00395 moles / 0.00198 moles = 2

For 1 mol N we have 2.33 moles C, 1.66 moles H and 2 moles O

OR

For 3 moles N we have 7 moles C, 5 moles H and 6 moles O

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Which water feature is often associated with condensation?
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Answer:

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3 years ago
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Which statements about spontaneous processes are true? Select all that apply: A spontaneous process is one that occurs very quic
Mkey [24]

Answer: Here are the complete options.

A spontaneous process is one that occurs very quickly. A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium. A spontaneous process is one that occurs without continuous input of energy from outside the system. A process is spontaneous if it must be continuously forced or driven.

The correct option is

A spontaneous process is one that occurs without continuous input of energy from outside the system.

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Explanation:

spontaneous process is one that occurs without continuous input of energy from outside the system and occur on its own because spontaneous processes are thermodynamically favorable characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy. Which means that the initial energy is higher than the final energy.

A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium which will result to The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative

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What metal has the largest atomic radius in period 4 (78 points)
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Answer:

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