Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases.
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.
1.
Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar
2. For the volume, let's find the total volume first.
V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
The volume of the new solution is
calculation
by use of the formula
M1V1= M2V2
M1 (molarity 1) = 2.13 M
V1 (volume 1) = 1.24 l
M2 ( molarity 2) = 1.60 M
V2 (volume 2) = ?
by making V2 the subject of the formula
V2 = M1 V1/ M2
V2=( 1.24 x 2.13)÷ 1.60= 1.651 L
