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djverab [1.8K]
1 year ago
6

A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field. What color migh

t MQ₆²⁺, be expected to show? Explain.
Chemistry
1 answer:
Bingel [31]1 year ago
4 0

A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field.  MQ₆²⁺, be expected to show green color.

<h3>What is spectrochemical series?</h3>

The ligands (attachment to a metal ion) are listed in the spectrochemical series according to the strength of their field. The series has been created by superimposing several sequences discovered through spectroscopic research because it is impossible to produce the full series by examining complexes with a single metal ion. The halides are referred to be weak-field ligands whereas the ligands cyanide and CO are strong-field ligands. Medium field effects are claimed to be produced by ligands like water and ammonia.

To know more about the spectrochemical series, visit:

brainly.com/question/27892620

#SPJ4

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The molar concentration (M) of a solution prepared by dissolving 0.2362g of Cr(NO3)3 in a 50-mL volumetric flask is 0.01985M, wh
Vinil7 [7]

Answer:

Here's what I get  

Explanation:

You want to dilute the original solution by a factor of 25 in two steps, so you could dilute it by a factor of 5 in the first step, then dilute the new solution by another factor of 5.

A. First dilution

Use a 10 mL pipet to transfer 10 mL of the original solution to a 50 mL volumetric flask. Make up to the mark with distilled water. Shake well to mix.

Use the dilution formula to calculate the new concentration.

\begin{array}{rcl}c_{1}V_{1} & = & c_{2}V_{2}\\0.01985 \times 10.00 & = & c_{2} \times 50.00\\0.1985 & = & 50.00 c_{2}\\\\c_{2}& = & \dfrac{0.1985}{50.00}\\\\& = & \text{0.003 970 mol/L}\\\end{array}

B. Second dilution

Repeat Step 1, using the 0.003 970 mol·L⁻¹ solution.

\begin{array}{rcl}c_{2}V_{2} & = & c_{3}V_{3}\\0.003970 \times 10.00 & = & c_{3} \times 50.00\\0.03970 & = & 50.00 c_{3}\\\\c_{3}& = & \dfrac{0.03970}{50.00}\\\\& = & \textbf{0.000 7940 mol/L}\\\end{array}\\\text{The concentration of the final solution is $\boxed{\textbf{0.000 7940 mol/L}}$}

3. Check:

Compare the final concentration with the original

\begin{array}{rcl}\dfrac{ c_{3}}{ c_{1}} & = & \dfrac{0.0007940}{0.01985}\\& = & \mathbf{\dfrac{1}{25.00}}\\\end{array}\\\text{The concentration of the final solution is } \boxed{\mathbf{\dfrac{1}{25}}} \text{ that of the original solution}

7 0
3 years ago
Is this a correct electron config?
lina2011 [118]
Yes, hope this helped!
3 0
3 years ago
Which of the following formulas represents an empirical formula?
luda_lava [24]

Answer:

C is the only option that represents an empirical formula, as it can't be simplified further.

You can turn C8H18, for example: to C4H9, but you can't do the same with C2H5.

Let me know if this helps!

8 0
3 years ago
50.0kg of nitrogen and 10.0kg 0f hydrogen are mixed to produce ammonia . calculate the ammonia formed, identify the limiting rea
Nana76 [90]

Answer:

Nitrogen is limiting reactant and 30.4kg of ammonia can be formed

Explanation:

Nitrogen, N₂, reacts with hydrogen, H₂ to produce ammonia, NH₃, as follows:

2N₂ + 3H₂ → 2NH₃

<em>Where 2 moles nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia</em>

<em />

To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles and the mass of ammonia formed as follows:

<em>Moles N2 -Molar mass: 28g/mol-</em>

50000g * (1mol / 28g) = 1786 moles N2

<em>Moles H2 -Molar mass: 2g/mol-</em>

10000g * (1mol / 2g) = 5000 moles H2

For a complete reaction of 5000 moles H2 are needed:

5000 mol H2 * (2mol N2 / 3mol H2) = 3333 moles N2. As there are just 1786 moles, Nitrogen is limiting reactant

The moles of ammonia that can be produced are 1786 moles because 2mol N2 = 2moles NH3.

The mass of ammonia -Molar mass NH3: 17g/mol- is:

1786 moles NH3 * (17g / mol) = 30362g =

30.4kg of ammonia can be formed

3 0
3 years ago
What does a large standard deviations in a date set mean?
mylen [45]
C. A high standard deviation means that the average distance from the data points to the mean is high, which is what C says.
5 0
3 years ago
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