What’s the equation for a nuclear reaction in which actinium undergoes alpha decay?

Explanation:

The most stable isotope of actinium, Ac-227, undergoes alpha decay.

Actinium is element 89, so the unbalanced nuclear equation is

It is convenient to replace the question by an atomic symbol, , where x = the atomic number, y = the mass number, and Z = the symbol of the element .

Then your equation becomes

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

89 = x + 2, so x = 89 - 2 = 87

227 = y + 4, so y = 223

Element 87 is francium, so the nuclear equation becomes

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

wher is the rest of the question