The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
Answer:
0.19 g
Explanation:
Step 1: Given data
Volume of hydrogen at standard temperature and pressure (STP): 2.1 L
Step 2: Calculate the moles corresponding to 2.1 L of hydrogen at STP
At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.
2.1 L × 1 mol/22.4 L = 0.094 mol
Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen
The molar mass of hydrogen is 2.02 g/mol.
0.094 mol × 2.02 g/mol = 0.19 g
Answer: 2738
Explaining: you gotta add
Answer:
b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Explanation:
The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.
Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.
Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.
Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
