A likely application of a radioactive isotope with a short half-life such as Technetium-99 will be as a medical tracer. It will likely be used by a doctor to check the movement of substances within a person's body.
A radioactive isotope with such a long half-life like Rubidium-87 is likely used in the determination of the age of fossils and artifacts found by archaeologists.<span />
J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. ... Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus.
Answer:
El número atómico de cada uno de los átomos es 26
Explanation:
El número de masa es la suma de las masas del protón y el neutrón de un átomo.
El número atómico es el número de protones en el átomo.
Los parámetros dados son;
La suma del número másico de ambos átomos = 110
La suma de los neutrones = 58
Por lo tanto, sea el número de protones y neutrones en un isótopo = P₁ y N₁ y el número de protones y neutrones en el otro isótopo = P₂ y N₂
Tenemos;
P₁ + N₁ + P₂ + N₂ = 110
N₁ + N₂ = 58
Por lo tanto;
P₁ + P₂ = 110 - (N₁ + N₂)
P₁ + P₂ = 110 - 58 = 52
Dado que los isótopos son del mismo elemento, sus protones serán iguales, por lo tanto;
P₁ = P₂
P₁ + P₂ = P₁ + P₁ = 2 × P₁
P₁ + P₂ = 52
2 × P₁ = 52
P₁ = 52/2 = 26 = P₂
El número atómico de ambos átomos es el número de protones en el átomo que es 26.
El número atómico del elemento del átomo es 26
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.