g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

Stoichiometrically, 1 mole of reacts with 2 moles of for a complete neutralization reaction.

Recall that: mole =

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole requires 2 moles HCl, then 0.0041 mole will require:

    0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

it is a good conductor

explanation:

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.