When radium-226 (atomic number-88) decays by emitting an alpha particle it becomes
a. polonium-222
c. radium-222
b. polonium-224
d. radon-222

d. Radon-222

Explanation:

²²⁶₈₈Ra → ²²²₈₆Rn  + ⁴₂He

Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.  

Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.  

Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.  

Now looking in the periodic table Radon is the only element with atomic number 86.  

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :