How many grams are in 1.19x10^27 particles of cl2

Option D. 1.40x10^5g

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 particles. This equally means that 1mole of Cl2 contains 6.02x10^23 particles.

1 mole of Cl2 =2 x 35.5 = 71g.

Now, If 71g of Cl2 contains 6.02x10^23 particles,

Then Xg of Cl2 will contain 1.19x10^27 particles i.e

Xg of Cl2 = (71x1.19x10^27)/(6.02x10^23)

Xg of Cl2 = 1.40x10^5g

Therefore, 1.40x10^5g of Cl2 contain 1.19x10^27 particles

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation:

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.