Answer:
CN^- is a strong field ligand
Explanation:
The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).
Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.
Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.
D has a total of four significant figures.
Answer:
Explanation:
We have in this question the equilibrium
X ( g ) + Y ( g ) ⇆ Z ( g )
With the equilibrium contant Kp = pZ/(pX x pY)
The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional
pV = nRT ⇒ p = nRT/V and n/V is molarity.
Therefore we can calculate the reaction quotient Q
Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2
Since Qp is greater than Kp the system proceeds from right to left.
We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.
Answer:
Four moles of the cation
Explanation:
2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)
Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.
The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.
This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.
Answer:
feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6
Explanation:
Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.
Overall balance: 
Water: 
Using substitution method, we have:
= 220.77 kg/s
= 4.16 kg/s
The maximum production rate of solid crystal is 
= 10*4.16 = 416 kg/s
Around evaporator:
kg/s
Around the mixing point:
Solid crystal: 
Using the last two equations, we can obtain:
kg/s
The rate of supplying fresh feed to obtain the production rate is:
= 352.5/220.77 = 1.6
