Quantity (g) of pure mgso4 in 2.4 g of mgso4•7h2o

1.1724 g.

Explanation:

The molecular mass of 1.0 mol (MgSO₄.7H₂O) = 246.4 g/mol.

The molecular mass of pure MgSO₄ = 120.366 g/mol.

The molecular mass of 7(H₂O) = 7(18.0 g/mol) = 126.0 g/mol.

∴ The mass % of pure MgSO₄ = [(mass of pure MgSO₄)/(mass of MgSO₄.7H₂O)] x 100 = [(120.366 g/mol)/(246.4 g/mol)] x 100 = 48.85%.

∴ the quantity of pure MgSO₄ = (mass of MgSO₄.7H₂O)(% of MgSO₄/100) = (2.4 g)(48.85/100) = 1.1724 g.

the solar system used to have much dust particles option no d

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.