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vladimir2022 [97]
3 years ago
12

5,6-dimethyl-2-octyne

Chemistry
1 answer:
Natalija [7]3 years ago
5 0
See photo for drawing :)

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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
vitfil [10]
[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)
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3 years ago
Which of the following elements has the greatest metallic character?<br> Rb<br> Mo<br> Au<br> B
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Answer:

Au. fifififiof

Explanation:

djdjdjndsksidndndnendnnd

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3 years ago
Read 2 more answers
5.5 cm east + 3.5 cm west =
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2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
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If a flame is yellow is it considered dirty or clean?And to make it blue do you add or take away air?
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It is clean and is blue when you take away air hope i helped.
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3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
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