Phosphorus-32 has a half-life of 14.3 days. How many milligrams of phosphorus-32 remain after 57.2 days of you start with 6.3 mg of the isotope

[A] = 0.39mg

Explanation:

The radioactive decay of an atom follows the equation:

Ln[A] = -kt + ln[A]₀

Where [A] is actual amount of isotope after time t; k is decay constant and [A]₀ initial amount of isotope

Also, we can write k in therms of half-life:

k = ln2 / Half-life

Solving for k:

k = ln2 / 14.3days

k = 0.04847days⁻¹

Solving in the equation for [A]:

Ln[A] = -0.04847days⁻¹*57.2days + ln[6.3mg]

ln[A] = -0.93193

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.