What mass of silver nitrate, agno3, is required to prepare 800. g of 3.50% solution of agno3

28 g of AgNo3

Explanation:

800 g of 3.5% solution of AgNo3 is to be prepared.

3.5% AgNo3 solution contains 3.5g of AgNo3 in 100 g of solution.

If 3.5 g AgNo3 is present in 100 g of solution, then how many g is needed to make 800 g of solution?

       3.5 g AgNo3100 g solution

        X g AgNo3 800 g solution

    X g AgNo3 = 3.5 x 800/100

                             = 28 g

Hence, 28 g of AgNo3 will be required to prepare 800 g of 3.5% solution of AgNo3.

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.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3