Answer:
The correct answer is - alpha particle and positron.
Explanation:
In this question, it is given that, 38^88Sr decays to 34^84Kr, which means there is an atomic number decrease by 4, 38 to 34, and atomic mass decreases by 4 as well 88 to 84.
A decrease in the atomic mass is possible only when there is an emission of the alpha particle as an alpha particle is made of 2 protons and 2 neutrons. If an atom emits an alpha particle, there is a change in atomic number as it decreases by two, and its mass number decreases by four.
So after the emission of an alpha particle, the new atom would be
38^88Sr=> 36^84X => 34^84Kr
so there is also two positron emission that leads to decrease in atomic number by one with each emission:
38^88Sr=> 2^4He+ 36^84X => 36^84X + 2(1^0β+) => 34^84Kr
Positron decay is the conversion of a proton into a neutron with the emission of a positron that causes the atomic number is decreased by one, which causes a change in the elemental identity of the daughter isotope.
Answer:
3.50*10^-11 mol3 dm-9
Explanation:
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
Ag2C2O4 --> 2Ag+ + C2O4 2-
So Ksp = [Ag+]^2 * [C2O42-]
In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.
So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)
= 3.50*10^-11 mol3 dm-9
Answer:
The reaction will shift to the right
Explanation:
In the reaction:
NO(g) + NO₂(g) ⇌ N₂O₃(g)
Kp of reaction is:
= 0.03
<em>Where P represents the pressures in equilibrium</em>
Replacing in the kP formula the initial pressures:
= 0.001
As the <em>Reaction quotient (Q) </em>is less than kP, <em>the reaction will shift to the right </em>producing more N₂O₃ until Q = kP.
Answer:
what is the quetion.......
Explanation:
