How much energy is released by the decay of 3 grams of 230th in the following reaction 230 th - 226ra + "he (230 th = 229.9837 g/mol, 226ra = 225.9771 g/mol, "he = 4.008 g/mol) (10 pts.)

The energy released for the decay of 3 grams of 230-Thorium is

Explanation:

First we have to calculate the mass defect .

The equation for the alpha decay of thorium nucleus follows:

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

(Conversion factor: )

To calculate the energy released, we use Einstein equation, which is:

The energy released for 230 grams of decay of thorium is

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of =

Then, 3 grams of Th will release energy of =

Hence, the energy released for the decay of 3 grams of 230-Thorium is

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g