Answer: -
First Ionization energy IE 1 for element X = 801
Here X is told to be in the third period.
So principal quantum number 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.
The iupac name of the compound will be hex-3-yne or 3-hexyne. By marking number of carbon in the given compound, it was found that triple bond comes at third position and there are total 6 carbon in the compound. So it will have Hex as a prefix and as it contains triple bond so it will have yne as a suffix and as the triple bond is at third position, so it will be hex-3-yne or 3-hexyne.
Answer:
Solids
Explanation:
At standard room temperature and pressure, most of the elements are solids.
Just a few of the elements are liquids and gases at this temperature.
The periodic table is made up of metals, metalloids and non-metals. Most of these substances are actually solid.
Some non-metals are gaseous at standard room temperature.
Answer:
The answer to your question is:
Vol of NO2 = 11.19 L
Vol of O2 = 2.8 L
Explanation:
Data
N2O5 = 56 g
STP T = 0°C = 273°K
P = 1 atm
MW N2O5 = 216 g
Gases law = PV = nRT
Process
216 g of N2O5 ---------------- 1 mol
54 g ----------------- x
x = (54 x 1) / 216
x = 0.25 mol of N2O5
2 mol of N2O5 ----------------- 4 mol of NO2
0.25 mol ------------------ x
x = (0.25 x 4) / 2 = 0.5 mol of NO2
V = nRT/P
V = (0.5)(0.082)(273) / 1 = 11.19 L
2 mol of N2O5 ----------------- 1 O2
0.25 N2O5 ---------------------- x
x = (0.25 x 1) / 2 = 0.125 mol
Vol = (0.125)((0.082)(273) / 1 = 2.8 L
Janice’s yard was probably not exposed to as much sun and heat as the yard across the street
