How many grams of Na2SO4 are required to make 0.30 L of 0.500 M Na2SO4?

21.3 g of are required to make 0.30 L of 0.500 M .

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,

n = moles of solute

= volume of solution in L

moles of =

Now put all the given values in the formula of molality, we get

Therefore, 21.3 g of are required to make 0.30 L of 0.500 M .

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)