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)
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Answer:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Explanation:
Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).
The equation can be written as follow:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
The above equation can be balance as illustrated below:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O
There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O
There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Now, the equation is balanced.
Answer:
A
Explanation:
Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.
As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain
Explanation:
It is known that in a simple cubic unit cell the atoms are only present at the corner of the unit cell. This means that there are in total 8 atoms present in a simple cubic unit cell.
Therefore, in one simple cubic unit cell sharing of one atom is only .
Hence, the total number of atoms in a unit cell will be as follows.
= 1
Thus, we can conclude that there is 1 calcium atom present in each unit cell.