Answer:
1.02mol
Explanation:
Using the general gas equation below;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information provided in this question,
P = 2.0 atm
V = 11.4L
T = 273K
n = ?
Using PV = nRT
n = PV/RT
n = 2 × 11.4/ 0.0821 × 273
n = 22.8/22.41
n = 1.017
n = 1.02mol
Glucose is a simple sugar with the molecular formula C6H12O6. It is a carbohydrate.
Answer:
Many areas of the United States experience explosive population growth. <u>The more people that reside someplace, the more demand there is for water there.</u> Often these urban-growth <u>expansions are unplanned and place extraordinary stress on the water supply system, mainly on the groundwater.</u> <u>The stress often depletes groundwater supply, thereby causing wells to dry up.</u> Then water must be brought from somewhere else to support the local population.
Such situations have occurred all over the United States. For example, increased population growth in the southwestern United States has significantly lowered the water table 50 to 200 feet (depending on the area) since the 1940s. Managing urban growth, efforts to reduce water demand, conservation of the resource, and attempts to increase the water supply all address the problem of exceeding water resource limits.
<u>Human activities affect groundwater quality.</u>
<u>Here are some sources </u>and possible solutions to groundwater pollution:
<u>Agriculture</u>—Reduce usage of pesticides and fertilizers.
<u>Landfills</u>—Monitor for leakage and repair linings.
<u>Underground storage tanks</u>—Remove damaged and unused tanks.
<u>Household wastes</u>—Properly dispose of household hazardous waste.
<u>Septic tank leaks</u>—Properly maintain and repair tanks.
Explanation:
This came from the K12 learning course read this and the answer will be there. I underlined the important parts for the answer.
Answer:
reactive nonmetals since they have a full valence shell (that's why they're stable).
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.
