A balloon contains 3.20 L of helium. The pressure is reduced to 59.0 kPa, and the balloon expands to occupy a volume of 5.50 L. What was the initial pressure exerted on
the balloon?

101.4pKa

Explanation:

Given parameters:

Initial Volume of helium  = 3.20L

Final pressure of helium  = 59kPa

Final volume on helium gas = 5.5L

Unknown:

Initial pressure exerted on the balloon = ?

Solution;

To solve this pressure - volume problem, we simply apply the Boyle's law.

It states that "the volume of a fixed mass gas varies inversely as the pressure changes if the temperature is constant".

  Mathematically;

         P1 V1  = P2 V2

Now, insert parameters and solve;

         P1 x 3.2 = 59 x 5.5

          P1  = 101.4pKa

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.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.