<span>Oxidation or reaction with oxygen is a chemical property</span>
Answer:
5.03 Naming Compounds that contain Polyatomic Ions
Explanation:
Trust me.
Answer:
It's 10 cenimeters per second
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:
B, D, E, C, A
Explanation:
We have 5 blocks with their respective masses and volumes.
Block Mass Volume
A 65.14 kg 103.38 L
B 0.64 kg 100.64 L
C 4.08 kg 104.08 L
D 3.10 kg 103.10 L
E 3.53 kg 101.00 L
The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V
ρA = 65.14 kg / 103.38 L = 0.6301 kg/L
ρB = 0.64 kg / 100.64 L = 0.0064 kg/L
ρC = 4.08 kg / 104.08 L = 0.0392 kg/L
ρD = 3.10 kg / 103.10 L = 0.0301 kg/L
ρE = 3.53 kg / 101.00 L = 0.0350 kg/L
The order from least dense to most dense is B, D, E, C, A
