How many grams of oxygen are required for the combustion of 25.5 grams of magnesium? 2mg(s) + o²(g) -> 2mgo(s)

Step 1 : Convert grams of magnesium to moles using n = m/MM.

MM is found in periodic table if you don't know it yet.

So we get :     n(Mg) = 25.5 g  ÷ 24.305 g/mol. = 1.0491... mol.

Step 2: Look at the stoichiometric ratios in the formula. Mg : O = 2:1.

Therefore, the moles of oxygen is half the number of moles of Magnesium.

n(O2) = 0.5 x 1.0491... mol = 0.52458... mol

Step 3: Now convert moles of Oxygen to grams using m = n x MM. Hey presto you have your answer!

m(O2) = 0.52458... x 2(15.999) = 16.7856... grams.

To 3 significant  figures this is 16.8grams.

Let me know if you are confused about anything.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

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.