How many milliliters of 0.10 m pb(no3)2 are required to react with 75 ml of 0.20 m nai? pb(no3)2 + 2 nai --> pbi2 + 2 nano3?

The Ml of 0.10 Pb(NO3)2 that are required to react with 75 Ml of 0.20M Nai is 75 ml

calculation

Pb(NO3)2 +2Nai → Pbi2 + 2NaNo3

find the number of moles of Nai used

moles= molarity x volume in liters

volume in liters = 75/1000=0.075 moles

mole =0.20 x0.075 = 0.015 moles

by use of mole ratio between Pb(NO3)2 to Nai which is 1 :2 the moles of Pb(NO3)2 = 0.015 x1/2 =7.5 x10^-3 moles

volume (ml) )= moles/molarity x 1000

=volume = (7.5 x10^-3) / 0.10 = 75 Ml

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

zfoods? ok, pressure is used to keep the heat inside and to keep the food perfect.