Chemical because its mixing two chemicals which are substances
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:
0.0018 mol/min is the average rate of carbon dioxide gas production in moles per minute during the 10 minutes.
Explanation:
Pressure of carbon dioxide gas = P = 0.25 atm
Volume of carbon dioxide gas = V = 5.0 L
Moles of carbon dioxide gas = n
Temperature of the carbon dioxide gas,T = 550°C = 550+273 K = 823 K
Using an ideal gas equation :
n = 0.018 mol
In 10 minutes pressure of carbon dioxide gas reached to 0.25 atm
Average rate of production of moles of carbon dioxide gas per minute in duration of 10 minutes:
0.0018 mol/min is the average rate of carbon dioxide gas production in moles per minute during the 10 minutes.
Answer:
6.61 × 10∧-29 m³
Explanation:
Given data:
Atomic radius= 143 pm = 143 × 10∧-12 m
volume = ?
Formula:
r = a/2√2
143 × 10∧-12 m = a/ 2√2
a= 143 × 10∧-12 m × 2√2
a= 404.4 × 10∧-12 m
where a is edge length, so we can calculate the volume by using following formula:
volume= a³
V= (404.4 × 10∧-12 m)³
v= 6.61 × 10∧-29 m³
