The test for this is fairly simple.
We take a glowing match or splint near the gas sample, if the glow intensifies, oxygen is present.
If a lit splint or match goes out with a popping sound, this means that hydrogen is present.
You have to figure out a way to write the two unknown abundances in terms of one variable.
The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).
203(X) + 205(1-X) = 204.4
Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.
After that the higher percentage would be the most abundant.
203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6
x = 0.3
1-x = 0.7
Then the TI-205 would have the highest percentage and would be the most abundant.
Answer:
2.77 mol N
Explanation:
M(N2O) = 2*14 + 16 = 44 g/mol
61.0 g * 1 mol/44g = (61/44) mol N2O
N2O ---- 2N
1 mol 2 mol
(61/44) mol x mol
x = (61/44)*2/1 = 2.77 mol N
Answer:
4
Explanation:
The balanced equation for the reaction is as follows:
4Al + 3O2 --> 2(Al2O3)
