Explanation:
The most reactive metals are found on the left of the periodic table, in the blue column, known as the alkali metals. Their reactivity increases as we go down column (group) one. Reactive metals, when attached to less reactive metals, have the ability to prevent the less reactive metal from rusting.
Answer:
C
Explanation:
This is essentially one of the several safety measures in the chemical laboratory. This particular approach is one used in the case of fire eventualities.
A is wrong
This is because in the advent of a fire incident, it is necessary to evacuate the building as a whole. Meeting in the hallway is still within the building which is not the right thing to do when there’s a fire outbreak. Occupants are expected to leave the building immediately
B. Is also wrong. Taking time to pack your belongings might make you be caught in the inferno. It is expected that you leave the building at once
Answer:
Option C = same period.
Explanation:
All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell.
However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.
We know that to relate solutions of with the factors of molarity and volume, we can use the equation:
**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
is the left side of the equation. Then we have:
We can then solve for
:
and
or
We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
