The answer is: D.unstable nuclei emitting high-energy particles as they formed more stable compositions.
Those high-energy particles are alpha particles
, beta particles
, gamma radiation.
For example, the decay chain of ²³⁸U is called the uranium series.
Decay start with U-238 and ends with Pb-206. There are several alpha and beta minus decays.
Antoine Henri Becquerel (1852 – 1908) was a French physicist and the first person to discover evidence of radioactivity.
Becquerel wrapped fluorescing crystal (uranium salt potassium uranyl sulfate) in a cloth, along with the photographic plate and a copper Maltese cross.
Several days later, he discovered that a image of the cross appeared on the plate.
The uranium salt was emitting radiation.
Because of this discovery, Becquerel won a Nobel Prize for Physics in 1903, which he shared with Marie Curie and Pierre Curie.
Answer:
Updrafts characterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop. In a mature storm, updrafts are present alongside downdrafts caused by cooling and by falling precipitation.
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Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Answer:
The water phase with the smallest temperature increase when adding 10 kcal of heat is solid ice.
Explanation:
The rest of the statements are incorrect. The density of ice is lower than the density of water. The heat capacity of solid ice is greater almost twice the heat capacity of the liquid water. The heat capacity of vapors is less than heat capacity of liquid.
Answer:
Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.
Explanation:
Hello,
In this case, for the dilution process from concentrated 12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

Which can also be written in terms of concentrations and volumes:

Thus, solving for the initial volume or aliquot that must be taken from the 12-M HCl, we obtain:

It means that you must take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water for such preparation.
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