An experiment requires 0.38 M NH3(aq). The stockroom manager estimates that 15 L of the base is needed. What volume of 15 M NH3(aq) will be used to prepare this amount of 0.38 M base? Answer in units of mL.

480 mL

Explanation:

15L  of  0.38M of NH3(aq.) means that we require 15x0.38 moles of NH3.

this is to be provided by 15M x V.  

Where V is volume in Liters.

These are equal  

⇒15M x V = 15x 0.48  

Hence, Volume required V = 0.48L =480mL

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.