The grams of N2 that are required to produce 100.0 l of NH3 at STP
At stp 1moles = 22.4 l. what about 100.0 L of NH3
= 100 / 22.4 lx1 moles = 4.46 moles of NH3
write the reacting equation
N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2
mass = moles x molar mass
= 2.23moles x 28 g/mol = 62.4 grams
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.
Sorry I came a lil late,
The answer to your question is, 2.
Hope this helps! :)
The number of valence electrons of an element can be determined by the periodic table group (vertical column) in which the element is categorized
Answer: The final temperature of nickel and water is 
.
Explanation:
The given data is as follows.
Mass of water, m = 55.0 g,
Initial temp, 
,
Final temp, 
= ?,
Specific heat of water = 4.184 
,
Now, we will calculate the heat energy as follows.
q = 
= 
Also,
mass of Ni, m = 15.0 g,
Initial temperature, 
,
Final temperature, 
= ?
Specific heat of nickel = 0.444
Hence, we will calculate the heat energy as follows.
q =
= 
Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.
= -()
=
Thus, we can conclude that the final temperature of nickel and water is .
