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andreyandreev [35.5K]
3 years ago
6

How many neutrons are in Cesium-130 (130/50 Cs)

Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
4 0
 Best Answer:<span>  </span><span>Cross sections for formation of cesium and rubidium isotopes produced by bombardment of uranium with protons ranging in energy from 0.1 to 6.2 Bev were measured both radiochemically and mass spectrometrically. Independent yields were determined for Rb/sup 84/, Rb/sup 86/, Cs/sup 127/, Sc, su p 129/. Cs/sup 130/, Cs/sup 131/, Cs/sup 132/, Cs/sup 134/, Cs/sup 136/, and, at some e nergies, Rb/sup 83 and Cs/sup 135/. In addition, the independent yield of Ba/sup 131/ and the chain yields of Cs/sup 125/, Cs/sup 127/, Cs/sup 129/, La/sup 131/, Cs/sup 135/, Cs/sup 137/, ion cross sections of the Cs and Ba products on the neutron- excess side of stability decrease monotonically with increasing energy above 0.1 Bev, whereas the excitation functions for independent formation of the more neutron-deficient products in the Cs-Ba region and of Rb/sup 84/ and Rb/sup 86/ all go through maxima. The proton energies at which these maxima occur fall on a smooth curve when plotted against the neutronproton ratio of the product, with the peaks moving to higher energies with decreasing neutron-proton ratio. Under the assumption that the mass-yield curve in the region 125 < A < 140 is rather flat at each proton energy, the crosssection data in the Cs region can be used to deduce the charge dispersion in this mass range. Plots of log sigma vs N/Z (or Z--Z/sub A/) show symmetrical bell-shaped peaks up to a bombarding energy of 0.38 Bev, with full width at halfmaximum increasing from 3.3 Z units at 0.10 Bev to about 5 Z units at 0.38 Bev, and with the peak position (Z/sub p/) moving from Z/ sub A/ -- 1.44 to Z/sub A/ -- 0.85 over the same energy range. At all higher energies, a double-peaked charge distribution was found, with a neutron-excess peak centered at N/Z approximates 1,515(Z/sub p/ approximates Z/sub A/ -- 1.9), and having approximately constant width and height at bombarding energies greater than 1 Bev. The peak on the neutron-deficient side which first becomes noticeable at 0.68 Bev appears to become broader and shift slightiy to smaller N/ Z values with increasing energy, The two peaks are of comparable height in the Bev region, and the peak-to-valley ratio is only approximates 2. The total formation cross section per mass number in the Cs region decreases from approximates 52 mb at 0.1 Bev to about 29 mb at 1 Bev and then stays approximately constant; the contribution of the neutron-excess peak above 1 Bev is about 12 mb. The neutron-excess peak corresponds in width and position to that obtained in fission by approximates 50-Mev protons. The recoil behavior of Ba/sup 140/ lends support to the idea that the neutron-excess products are formed in a lowdeposition-energy process. The recoil behavior of Ba/sup 131/ indicates that it is formed in a high-deposition-energy process. Post-fission neutron evaporation is required for the observed characteristics of the excitation functions of the rubidium isotopes and the neutron-deficient species in the Cs region. The correlation between neutron-proton ratios and positions of excitation function maxima is semiquantitatively accounted for if fission with unchanged charge distribution, followed by nucleon evaporation, is assumed. (auth) 
</span>
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Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a
Ghella [55]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image  

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule  is gotten according to longest carbon chain  = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

3 0
3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
3 years ago
A molecule with the formula C18H36O2 is probably a?
Alina [70]

Ethyl palmitate is an organic compound with the chemical formula C18H36O2. It is a colorless solid with a wax-like odor. Chemically, ethyl palmitate is the ethyl ester of palmitic acid.

Ethyl palmitate is used as a hair- and skin-conditioning agent.<span />


3 0
3 years ago
Read 2 more answers
Which of the following chemical formula show an iconic compound
BaLLatris [955]
This question comes with this list of answer choices:

<span>A. CO₂

B.N₂O

C.NaCl

</span><span> D.C₆H₁₂O₆
</span>

Answer: option C.NaCl

Justification.

1) When two diferent elements join together chemically they can form either a covalent bond or a ionic bond.

2) When the bonds are covalents the compound is called covalent, when the bonds are ionics the compound is called ionic.

3) To predict whether the compound is ionic or covalent, you must see the electronegativities of the elements.

4) Ionic compounds require the formation of ions, a posititve ion and a negative ion. The positive ion is formed when the element loses one or more electrons. The negative ion is formed when the element gains one or more electrons.

Then, in the formation of a ionic compound it is implicit the exchange of electrons (one element gains the electrons that the other lose) and this is possible, when the difference in electronegativities are high enough.

5) So, you must look at the electronegativities to make your prediction. In the given options, the compounds of the options A, B and D are between non-metals. Then, you can predict that their electronegativities are not as much differet as to form ions. They will form covalent bonds.


6) Being sodium and chloride a metal and non-metal, you can expect that their electronegativities differ quite enough to form the ionic bond (this is not true for any pair of non-metal and metal elements, but being them the only pair of metal - non metal elements, this is the only option).

This leads to the conclusion that the NaCl formula shows an ionic compound, and is the answer.

5 0
3 years ago
How do I do 1-2? Do I use graph paper?
sergejj [24]
This would equal -1 or negative one
5 0
3 years ago
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