Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
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
Answer:
T₂ = 392 K
Explanation:
Given that,
Initial volume of the hot air balloon, V₁ = 55500 m³
Initial temperature, T₁ = 21°C = 294 K
Final volume, V₂ = 74000 m³
We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.
Where
T₂ is the final temperature
So,
So, the new temperature is 392 K.
Answer:
energy which a body possesses by virtue of being in motion
Explanation:
the work needed ro accelerate a body of given mass
Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>
