<span>You are given O2 and C3H8, this is a combustion
reaction. The chemical reaction is C3H8 + 10O2 à 3CO2 + 4H2O. You are also given the molar mass
of O2 which is 32.00 g/mol and C3H8 which is 44.1 g/mol. You are required to
find the mass of O2 in grams. Since you have the reaction, oe mole of C3H8 is
required to completely react 10 moles of O2. So,</span>
0.025g C3H8(1 mol C3H8/44.1 g C3H8)(10 mol O2/1
mol C3H8)(32 g O2/1 mol O2) = <u>0.1802 g O2
</u>
<span> </span>
Crystal field splitting is the difference in energy between d orbitals of ligands. Crystal field splitting number is denoted by the capital Greek letter Δ. Crystal field splitting explains the difference in color between two similar metal-ligand complexes.
<h3>What is crystal field splitting of d-orbitals?</h3>
The splitting of fivefold degenerate d orbitals of the metal ion into two levels in a tetrahedral crystal field is the representation of two sets of orbitals as Td. The electrons in dx2-y2 and dz2 orbitals are less repelled by the ligands than the electrons present in dxy, dyz, and dxz orbitals.
<h3>Which of the following factors affect crystal field splitting energy?</h3>
There are the following factors that affect the crystal field splitting. These are the nature of ligands, coordination number, arrangement of ligand, size of a metal atom, charge on the metal atom, size of ligands, electronegativity, and interatomic distance.
Learn more about crystal field splitting here:
<h3>
brainly.com/question/13004475</h3>
<h3>#SPJ4</h3>
the final concentration of NaI solution in 60 grams/litre.
Explanation:
Given that:
Initial concentration of NaI solution M1 = 0.2 M
initial volume of NaI V1 = 2 L
Final volume V2 = 1 Litre
Final molarity=?
concentration in grams/litre = ?
molar mass of NaI = 150 gram/mole
For dilution following formula is used:
M1 V1 = M2V2
putting the values in the equation
0.2 X 2 = 1 X M2
M2 = 0.4
For concentration in grams/litre, formula used
molarity =
mass = 0.4 x 150
= 60 grams
So, 60 grams of NaI will be present in final solution of NaI after evaporation.
The concentration is 60 grams/ L (as volume got reduced to 1 litre from 2 litres)
Answer:
it has 9 sulfur atoms
Explanation:
you have to multiply the 3 outside the bracket by the one in side to get your answer