Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
Learn more about mole fraction:
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It is called a waxxing gibbous, pls brainliest
Theoretical yield is the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. It is the amount of product resulting from a perfect chemical reaction and thus not the same as the amount you'll actually get from a reaction.
Answer:
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not.
A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons.
Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it.
So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’.
You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not.
Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration
Co3+:1s22s22p63s23p63d6
For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split.
More specifically, you can say that
a strong field ligand will produce a more significant splitting energy, Δ a weak field ligand will produce a less significant splitting energy, Δ
Now, the spectrochemical series looks like this
http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm
Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands.
http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media...
In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed.
This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic.
On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed.
Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.
