Jack has 100 milliliters of a 12 molar solution of sulfuric acid. how much of it should he put into a graduated cylinder to make 20 milliliters of a 1.2 molar solution? i'm studying for asvab can someone walk me through the solution

This is a dilution equation where a certain volume from a more concentrated solution is diluted with distilled water to make a less concentrated solution. We can use the dilution equation as follows

c1v1 = c2v2

where c1 is concentration and v1 is volume of the more concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

where c1 - 12 M

c2 - 1.2 M

v2 - 20 mL

we need to find v1 , that is the volume that should be taken from the 12 M solution to make 20 mL of 1.2 M solution

substituting these values in the equation

12 M x v1 = 1.2 M x 20 mL

v1 = 2 mL

therefore 2 mL of 12 M solution should be taken to the graduated cylinder and diluted up to 20 mL to make a 1.2 M solution

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.