1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Svetach [21]
3 years ago
11

The three beakers shown below contain solutions of [cof6]3–, [co(nh3)6]3+, and [co(cn)6]3–. based on the colors of the three sol

utions, which compound is present in each of the beakers?
Chemistry
1 answer:
vitfil [10]3 years ago
3 0
[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.

- Ligands strength order:     CN⁻ > NH₃ > F⁻ 

- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻

- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.

- So CN complex will absorb at lower wavelength (yellow color)
You might be interested in
How many formula units are in 239.2g of <br> Br2<br> MgCl2<br> H2O<br> Fe
Oksana_A [137]

The formula units in the substances are as follows:

  • Br2 = 8.99 × 10^23 formula units
  • MgCl2 = 1.51 × 10^24 formula units
  • H2O = 2.57 × 10^24 formula units
  • Fe = 2.57 × 10^24 formula units

<h3>How many moles are in 239.2 g of the given substances?</h3>

The moles of the substances are determined from their molar mass.

Molar mass of the substances is given as follows:

  • Br2 = 160 g/mol
  • MgCl2 = 95 g/mol
  • H2O = 18 g/mol
  • Fe = 56 g/mol

Formula units = mass/molar mass × 6.02 × 10^23

The formula units in the substances are as follows:

  • Br2 = 239.2/160 × 6.02 × 10^23 = 8.99 × 10^23 formula units
  • MgCl2 = 239.2/95 × 6.02 × 10^23 = 1.51 × 10^24 formula units
  • H2O = 239.2/18 × 6.02 × 10^23 = 2.57 × 10^24 formula units
  • Fe = 239.2/56 × 6.02 × 10^23 = 2.57 × 10^24 formula units

In conclusion, the number of formula units is derived from the moles and Avogadro number.

Learn more about formula units at: brainly.com/question/24529075

#SPJ1

6 0
2 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
3 years ago
The following compound has been found effective in treating pain and inflammation (J. Med Chem. 2007, 4222). Which sequence corr
love history [14]
<h3><u>Full Question:</u></h3>

The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?

A) 1 < 2 < 3

B) 2 < 3 < 1

C) 3 < 1 < 2

D) 1 < 3 < 2

<h3><u>Answer: </u></h3>

The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.

Option B

<h3><u>Explanation. </u></h3>

Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.

A carbonyl compound is a sp^2 hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.

In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.

7 0
3 years ago
what is the products of the reaction between 2-methylbut-2-ene and HCL 1b. write the IUPAC names of the product of the reaction;
zzz [600]

Answer:

Explanation:

2 - chloro-2-methylbutane

8 0
2 years ago
Why do you think we are able to study the atom in the modern world?
andriy [413]
Advancement in technology to help with detailed studies
6 0
2 years ago
Other questions:
  • When carbon undergoes sp2 hybridization it forms
    12·1 answer
  • Onetaq standard reaction buffer fungal metagenomics
    15·1 answer
  • The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consist
    7·2 answers
  • Writing the net ionic equation
    14·2 answers
  • The total resistance in this circuit is
    9·1 answer
  • How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through
    13·1 answer
  • What type of energy relates to the movement of objects or its position in gravity ?
    9·1 answer
  • The average kinetic energy of water at 25oC is
    8·1 answer
  • ILL BRAINLIST ITS URGENTTT!!!!
    12·1 answer
  • Be sure to answer all parts. an aqueous solution of a strong base has a ph of 10. 220 at 25°c. calculate the concentration of th
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!