The amount of i−3(aq) in a solution can be determined by titration with a solution containing a known concentration of s2o2−3(aq) (thiosulfate ion). the determination is based on the net ionic equation 2s2o2−3(aq)+i3(aq)⟶s4o2−6(aq)+3i−(a q) given that it requires 29.6 ml of 0.260 m na2s2o3(aq) to titrate a 30.0 ml sample of i−3(aq), calculate the molarity of i−3(aq) in the solution.

Answer : The molarity of in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

First we have to calculate the moles of .

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of .

From the balanced chemical reaction, we conclude that

As, 2 moles of react with 1 mole of

So, 0.007696 moles of react with mole of

The moles of = 0.003848 mole

Now we have to calculate the molarity of .

Now put all the given values in this formula, we get:

Therefore, the molarity of in the solution is, 0.128 M

chemistry sucks a**

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.