Answer:
The answer to your question would be "the insect touching the trigger hairs".
Explanation:
I'm not much of an ex-plainer but I know this is the right answer because I took the test and got a 100%. Please trust me on this. If wrong, please tell me. This is what I was taught at school. Thank you and good day.
Oregon trail would be the best answer
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.
Answer: 353.6 g.
Explanation:
1) Word equation (given)
Sodium phosphate + calcium choride → sodium chloride + calcium phosphate
2) Balanced chemical equation
2 Na₃PO₄ +3 CaCl₂ → 6 NaCl + Ca₃(PO4)₂
3) Mole ratio
2 moles Na₃PO₄ : 3 moles CaCl₂ : 6 moles NaCl : 1 mole Ca₃(PO4)₂
4) Formula to convert grams to moles
number of moles = mass in grams / molar mass
5) Number of moles of limiting reactant
a) mass in grams = 379.4 g CaCl₂
b) molar mass CaCl₂ = 110.98 g/mol
c) number of moles CaCl₂ = 379.4 g / 110.98 g/mol = 3.419 moles
6) Theoretical yield of Ca₃(PO4)₂
3 moles CaCl₂ / 1 mole Ca₃(PO4)₂ = 3.419 moles CaCl₂ / x ⇒
x = 3.419 × 1 / 3 = 1.140 moles Ca₃(PO4)₂
7) Convert 1.140 moles Ca₃(PO4)₂
a) molar mass Ca₃(PO4)₂ = 310.1767 g/mol
b) mass in grams = number of moles × molar mass = 1.140 moles × 310.1767 g/mol = 353.6 g ← answer.
Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old
