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Ann [662]
3 years ago
6

2. Among the complexes K4[Cr(CN)6] (A),K4[Fe(CN)6] (B) , K3[Co(CN)6] (C) , K4[Mn(CN)6] (D), Jahn-Teller distortion is shown by *

D (0/1 Points) A,B and C B,C and D A and D B and C
Chemistry
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

K4[Cr(CN)6] and K4[Mn(CN)6]

Explanation:

The Jahn–Teller effect, sometimes also known as Jahn–Teller distortion, describes the geometrical distortion of molecules and ions that is associated with certain electron configurations. In Jahn Teller effect, more electrons occupy the lower orbital than the upper one, resulting in an overall lowering of the electronic energy. This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds. In octahedral complexes, the Jahn–Teller effect is most pronounced when an odd number of electrons occupy the eg orbitals. This situation is most pronounced in complexes with the configurations d9 , low-spin d7 or high-spin d4. These configurations display a strong Jahn Teller effect. Jahn Teller may also be observed if there is a degeneracy due to the electrons in the t2g orbitals. A weak Jahn Teller effect is also observed for d5 species.

K4[Cr(CN)6] contains Cr^2+ which is a d4 specie, hence it exhibits Jahn Teller effect. Also, K4[Mn(CN)6] contains Mn^2+ which is a d5 specie, therefore it also exhibits Jahn Teller effect.

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Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess
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Answer:

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

Explanation:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The relation of K_c\& K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p= Equilibrium constant in terms of partial pressure.=98.1

K_c= Equilibrium constant in terms of concentration  =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

\Delta n_g = Difference between gaseous moles on product side and reactant side=n_{g,p}-n_{g.r}=1-2=-1

98.1=K_c(RT)^{-1}

98.1 =\frac{K_c}{RT}

K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

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3 years ago
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Which element would make a poor, brittle conductor? use the periodic table for help. a. sulfur b. boron c. cobalt d. calcium?
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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

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Answer:

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