Neon = app. 20.1795 g/mol
<span>Bromine = app. 79.904 g/mol </span>
<span>-----> 79.904 g/mol / 20.1795 g/mol = 3.96 (Close to 4) </span>
<span>Using Moles In 1000 g </span>
<span>1000 g / 20.1795 g/mol = app. 49.555 mol of Ne </span>
<span>1000 g / 79.904 g/mol = app. 12.515 mol of Br </span>
<span>-----> 49.555 mol / 12.515 mol = 3.96 (Close to 4) </span>
<span>Using Avogadro's Number </span>
<span>49.555 mol x 6.022x10^23 atoms = app. 2.984x10^25 atoms of Ne </span>
<span>12.515 mol x 6.022x10^23 atoms = app. 7.537x10^24 atoms of Br </span>
<span>-----> 2.984x10^25 / 7.537x10^24 = 3.96 (Close to 4) </span>
<span>So no matter how you look at it or calculate it, the answer is always the same. </span>
<span>I hope it helps!</span>
Answer:
This a simple stoichiometry problem using the ideal gas law.
First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol
Now use a molar conversion using the balanced equation,
1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas
PV=nRT
V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L
The given question is incomplete. The complete question is:
Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.
Answer: 0.0484 g
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 0.920 A
t= time in seconds = 47.0 sec
96500 Coloumb of electricity electrolyzes 1 mole of Ag
43.24 C of electricity deposits =
of Ag
Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g
If it is just ratio you must just put the mass number on the either side of the symbol ‘:’ and cancel but the same process is not used to find simplest ratio
