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.
Answer:
a) 38.2 % mass
b) 61.8 g solute/100 g solvent
c) 1.65 g/mL
Explanation:
Given the data:
mass of solute = 17.5 g
mass of solvent= 28.3 g
total solution volume= 27.8 mL
a)- mass percent= mass of solute/mass of solution x 100
mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g
mass % = 17.5 g/45.8 g x 100 = 38.2 % mass
b)- solubility = grams of solute/ 100 g solvent
= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent
c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL
Answer is: D. It is not sodium bicarbonate.
Balanced chemical reaction of heating sodium bicarbonate: 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.
This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.
Answer:
34.2 g is the mass of carbon dioxide gas one have in the container.
Explanation:
Moles of 
:-
Mass = 49.8 g
Molar mass of oxygen gas = 32 g/mol
The formula for the calculation of moles is shown below:
Thus,
Since pressure and volume are constant, we can use the Avogadro's law as:-
Given ,
V₂ is twice the volume of V₁
V₂ = 2V₁
n₁ = ?
n₂ = 1.55625 mol
Using above equation as:
n₁ = 0.778125 moles
Moles of carbon dioxide = 0.778125 moles
Molar mass of 
= 44.0 g/mol
Mass of = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g
<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
