P: ?
V: 35,5L = 35,5 dm³
n: 0,54 mol
R: 83,14 hPa·dm³/mol·K
T: 223K
..................
pV = nRT
p = nRT/V
p = (0,54×83,14×223)/35,5
p = 282,02 hPa
Both transition metals and alkali metals are good conductors of heat and electricity, react with water, and are easily oxidized.
<h3>What are alkali metals and transition metals?</h3>
The alkali metals are elements of group 1 which are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). They are also known as the s-block elements because they have their outermost electron in an s-orbital.
The alkali metals are shiny, soft, highly reactive metals and readily lose their outermost electron to create cations with charge +1. They can tarnish rapidly in the air due to oxidation by atmospheric moisture and oxygen.
Transition elements or transition metals are elements that have partially filled d-orbitals. An element having a d-subshell that is partially filled with electrons or can form stable cations with an incompletely filled d orbital.
Any element present in the d-block of the modern periodic table which consists of groups 3 to 12, is considered to be a transition element. For example, the mercury in the +2 oxidation state, corresponds to an electronic configuration of (n-1)d¹⁰. Many paramagnetic compounds are formed by transition metals because they have unpaired electrons in the d orbital.
Learn more about transition metals and alkali metals, here:
brainly.com/question/15775417
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Answer:
A
Explanation:
A: Metals are ofthen the best conductors.
B: Metals and non-metals can be smooth or rough
C: Metals and non-metals can be hard to the touch
D: Metals and non-metals can float in water
2Na^+ and S^2-
(sodium cation and sulfide anion)
Answer:
2.16 MeV
Explanation:
To determine the amount of work done that is needed to assemble the atomic mass; we need to apply the equation;
U = 
where:
= proportionality constant = 
e = magnitude of the charge of each electron = 
r = length of each side of the vertex = 
So; replacing our values into above equation; we have:
U = 
U = 3.456 × 10 ⁻¹³ J
If we have to convert our unit from J to Mev; then we are going to have:
U = 3.456 × 10 ⁻¹³ J 
U = 2.16 MeV
Therefore, the amount of work done needed to assemble an atomic nucleus = 2.16 MeV
