Answer:
3.711 L
Explanation:
The formula you need to use is the following:
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C)
3.4L / 298 K = V2 / 273 K
V2 = 3.711 L
Answer:
Barium has the same number of valence electrons as calcium
Explanation:
Valence electrons is the number of electrons of an atom on the outer shell.
Those valence electrons can participate in the formation of a chemical bond (if the outer shell is not closed); in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
<u>Calcium</u> is an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons.
<u>Sulfur </u>is part of a group 16, called the chalcogens or oxygen family. Those atoms have 6 valence electrons. They can form a bound with atoms of group 2 such as calcium, but do not have the same number of valence electrons.
<u>Potassium</u> is part of group 1, called the alkali metals or lithium family. Those atoms have 1 valence electrons. That means Potassium do not have the same number of valence electrons like calcium.
<u>Neon</u> is part of group 18, the noble gasses. Those are stable atoms, which means they have 8 valence electrons. They do not have the same number of valence electrons like Calcium.
<u>Barium</u> an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons. Calcium is also part of this group.
This means barium has the same number of valence electrons as Calcium.
The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The dimensionless standard atomic weight instead refers to the AVERAGE of atomic mass values of a typical naturally-occurring mixture of isotopes for a sample of an element.
You can count it by yourself using formula
m = ({first isotopic distribution%}× {first atomic.mass})+ ({second isotopic distribution%}× {second atomic.mass}) / {100}