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Shalnov [3]
3 years ago
8

A storm is

Chemistry
2 answers:
IceJOKER [234]3 years ago
8 0

Answer:

I think it is B I'm not Sure

STALIN [3.7K]3 years ago
4 0
The answer is D. because A is a storm surge, B. Is air mass, and C is an occluded front (i think). i hope i’m correct i’m pretty sure i am.
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M → M+ + e- reduction or oxidation?
Julli [10]
That would be oxidation.

Oxidation is loss of electrons
Reduction is gain of electrons
4 0
3 years ago
How many neutrons are in Cesium-130 (130/50 Cs)
Sergeeva-Olga [200]
 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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4 0
3 years ago
This law brings Boyle's, Charle's, Avogadro's, and Gay-Lussac's laws together in one neat, easy to use package: PV = nRT. Using
NikAS [45]

Answer:

22.4L

Explanation:

The following data were obtained from the question:

P = 1atm

T = 273K

n = 1mole

R = 0.08206 L∙atm/mol∙K

V =?

PV = nRT

V = nRT/P

V = (1 x 0.08206 x 273)/1

V = 22.4L

8 0
3 years ago
Explain in at least five sentences what you’ve learned about how global warming and climate change are impacting our Earth in di
pochemuha

Answer:

a gradual increase in the overall temperature of the earth's atmosphere generally attributed to the greenhouse effect caused by increased levels of carbon dioxide, chlorofluorocarbons, and other pollutants

Explanation:

7 0
3 years ago
Calculate the heat (in calories) released when 115 g of CHCl3 changes from a liquid to a solid at
Alexandra [31]

Answer:  8855 Joules energy is released when 115 g of CHCl_3 changes from a liquid to a solid at  its freezing point.

Explanation:

Latent heat of freezing is the amount of heat released to convert 1 mole of  liquid to solid at atmospheric pressure.

Amount of heat released to freeze 1 gram of CHCl_3 = 77 J/g

Mass of CHCl_3 given = 115 gram

Heat released when 1 g of CHCl_3 is freezed = 77 J

Thus Heat released when 115 g of CHCl_3 is freezed =\frac{77}{1}\times 115=8855J

Thus 8855 Joules energy is released when 115 g of CHCl_3 changes from a liquid to a solid at  its freezing point.

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3 years ago
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