Hey there!
Atomic mass Bromine ( Br ) = 79.9 u
Therefore:
79.9 g Br --------------- 22.4 L ( at STP )
18.0 g Br --------------- volume ??
Volume Br = 18.0 * 22.4 / 79.9
Volume Br = 403.2 / 79.9
Volume Br = 5.046 L
hope this helps!
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
Hahahaaaa none of the above but IF <span>(c) is
1/2 mole of NaCl and 1/3 mole of MgCl2 instead,
then C is the right ans :)</span>
Answer:
i could help you can you explain more plzzzzzz i really need points
Explanation:
2Mg+ O₂= 2MgO
The charge of Mg is+2 and the charge of Oxygen is -2. 2 oxygen atoms will have a charge of -4 so the Mg atoms have to equal +4. A coefficient of 2 for the Mg atoms will balance it out
