<p>Question #46 from the 1994 Chem Subject test.</p>
<p>What is the empirical formula of a compound that contains .025 mole of Cd, .050 mole of C, and .100 mole of O?</p>
<p>(A) CdCO
(B) CdCO2
(C) CdCO3
(D) CdC2O2
(E) CdC2O4</p>
<p>The right answer is E
I wish I was better with numbers</p>
<p>Proportions, largest over smallest.</p>
<p>For O: .1/.025= 4
For C: .05/.025=2
And Cd: .025/.025=1</p>
<p>Twice as many Oxygen as Carbon, twice as many carbon as Cd</p>
<p>CdC2O4</p>
<p>:)</p>
<p>Wow. That seems so simple now. Mind if you simplify another one? :)</p>
<p>AlCl3 +NH3 + H20 ----->
Aluminium (III) chloride +ammonia +water----></p>
<p>Which of the following is one of the products obtained from the reaction above?
(A) AlN
(B) AlH3
(C) Al
(D) Al(NO3)3
(E) Al(OH)3</p>
<p>Right answer is (E) again. Thanks!</p>
<p>It is pretty much just a double replacement where NH3 + H20 turns into NH4OH. Then you just find OH. :D</p>
<p>I’m fairly sure it’s because aluminum hydroxide is insoluble, and thus precipitates out of solution. It’s the only real chemical change in the reaction, everything else is just spectators.</p>
<p>yes, that’s right because aluminum is a metal so it wants to form a base. thanks!</p>