Answer:
1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.
2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.
The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.
This are some of the transuranium elements:
Neptunio - 93
Plutonium - 94
Americium - 95
Curium - 96
Berkelium - 97
Californium - 98
Einstenium - 99
And so all the known elements (the last one is the 118).
3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.
4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:
0
β, which means 0 mass and charge - 1.
-1
5) This is, then, an example of a β decay equation for one transuranium element:
239 239 0
Np → Pu + β
93 94 -1
As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.
Explanation:
For the atom to be neutral, it must have the same number of electrons and protons. The aromic number indicates the number of protons. Since Carbon has 6 protons, it would have to have 6 electrons for it to be neutral
Answer:
2NO(g) + O2(g) --> 2NO2(g)
now 400 ml of NO × 2 mol of NO2/2 mol of NO
= 400 ml of NO2
now 500 ml of O2 × 2 mol of NO2/1 mol of O2
= 1000 ml of NO2
now 400 ml of NO2 × 1 mol of O2/2 mol of NO
= 200 ml
subtract that from 500 ml of total i.e. 500-200 =300 ml
The total volume of the reaction mixture is 1000 ml -300ml = 700 ml
Answer:
A.) ![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Explanation:
The general Kb expression is:
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
They have the same number of protons