This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
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Answer:
the answer is C + H2O # CO + H2.
Answer:
1, 3, 2
Explanation:
N2 + H2 → NH3
I usually find that the best way to systematically balance an equation by inspection is to start with the most complicated-looking formula and then balance atoms in the order:
- All atoms other than O and H
- O
- H
(a) The most complicated formula is NH3.
(b) Balance N.
We have 1 H in NH3, but 2 N on the left. We need 2 N on the right. Put a 1 in front of N2 and a 2 in front of NH3.
1N2 + H2 → 2NH3
(c) Balance H.
We have fixed 6 H on the right, so we need 6 H on the left. Put a 3 in front of H2.
1N2 + 3H2 → 2NH3
The equation is now balanced, and the coefficients are 1, 3, 2.
