Answer:
the bond b/w oxygen and hydrogen atom
What helps me to balance equations is to list the elements i have on each side of the equation, and use tally marks to see what I have and don't have. Then when I'm done balancing, I tally again to make sure everything matches up.
On the left side, you have 1 Al, and 2 O. On the right side, 1 Al and 3 O.
In order for the equation to balance, you need to place a 2 in front of the AlO on the right side. This would make the Al have 2 atoms and the O have six. On the left side, you need to place a 2 in front of the Al and a 3 in front of the O, making it six. Left side: 2 Al's 6 O's. Right side: 2 Al's and 6 O's. Matches!
Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
Answer:
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Explanation:
used in fire extinction, blasting rubber, foaming rubber and plastic
