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:
See explanation and image attached
Explanation:
Yttrium has many isotopes, the lowest mass number of Yttrium is 89Y.
Recall that electron capture converts an electron into a proton and then into a neutron with a consequent emission of a neutrino (v).
In electron capture, the mass number of the daughter nucleus remains the same as that of the parent nucleus while the atomic number of the daughter nucleus is less than that of the parent by one unit.
<span>To solve this exercise you need to know that to create CO₂ with C₂H₂ is necessary to have oxygen. So, the following balanced equation represents the reaction:
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)
Notice that 2 moles of C₂H₂ form 4 moles of </span><span>CO₂, so if </span>3.3 moles of C₂H₂ react, how many moles of CO2 would be produced?
2 moles <span>of C₂H₂ -------</span>4 moles of <span>CO₂
3.3 </span><span>moles <span>of C₂H₂--------x moles of CO₂
x=6.6 </span></span><span>moles of CO₂ produced.</span>
Answer:
37.5 L
Explanation:
Initial Volume, V1 = 15L
Initial Pressure P1 = 4.5 atm
Final Pressure, P2 = 1.8 atm
Final Volume V2 = ?
The relationship between these variables is given as;
P1V1 = P2V2
V2 = PIV1 / P1
Inserting the values;
V2 = 4.5 * 15 / 1.8
V2 = 37.5 L
