Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Explanation:
Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.
Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.
In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

Thus for mass number : 235 = 207+4X
4X= 28
X = 7
Thus for atomic number : 92 = 82+2X-Y
2X- Y = 10
2(7) - Y= 10
14-10 = Y
Y= 4

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Answer:
Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.
Answer:
The answer to your question is Argon
Explanation:
Electron configuration given 1s² 2s² 2p⁶ 3s² 3p⁶
To find the element whose electron configuration is given, we can do it by two methods.
Number 1. Sum all the exponents the result will give you the atomic number of the element.
2 + 2 + 6 + 2 + 6 = 18
The element with an atomic number of 18 is Argon.
Number 2. Look at the last terms of the electronic configuration
3s² 3p⁶
Number three indicates that this element is in the third period in the periodic table.
Sum the exponents 2 + 6 = 8
Number 8 indicates that this element is the number 8 of that period without considering the transition elements.
The element with these characteristics is Argon.
A. hydroxide is the answer
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very <u>strong base</u> (<em>sodium ethoxide</em>). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an <u>E2 mechanism</u>, therefore, the hydrogen that is removed must have an <u>angle of 180º</u> with respect to the leaving group (the "OH"). This is known as the <u>anti-periplanar configuration</u>.
The hydrogen that has this configuration is the one that placed with the <u>dashed bond</u> (<em>red hydrogen</em>). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!