Chem Test 唔該show steps~因為我唔知點計

1. Assume there are 100 grams of X2S.

So, mass of S in the compound

= 100 X (1 - 70.8%)

= 29.2 g

Molar mass of S = 32.1 g/mol

Therefore, no. of mole of S

= Mass / molar mass

= 29.2 / 32.1

= 0.910 mol

Molar ratio of X to S = 2 : 1

So, no. of mole of X

= 1.82 mol

Mass of X = 70.8 g

Therefore, molar mass of X

= Mass / no. of mole

= 70.8 / 1.82

= 39.0 g/mol

So, relative atomic mass of X is 39.0.

The answer is C.




2. Molar mass of oxygen atom

= 16.0 g/mol

So, no. of mole of oxygen

= mass / molar mass

= 16.0 / 16.0

= 1.0 mol

Molar ratio of X to oxygen = 2 : 1

So, the empirical formula is X2O.

So, the answer is D.

Therefore,

Let A be the relative atomic mass of X.

(AX2)/(AX2 + 32.1) X100% = 70.8%

A= 38.9157
so, the ans is C

2007-08-10 19:49:24 補充：
no. of mole of X=2no. of mole of O= 16/16=1SO, XO2 (B) is the ans