<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
The chemical reaction would be written as follows:
2Al + 3Cl2 = 2AlCl3
We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:
19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
Answer:
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Answer:
Make an observation.
Ask a question.
Form a hypothesis, or testable explanation.
Make a prediction based on the hypothesis.
Test the prediction.
Iterate: use the results to make new hypotheses or predictions.
Explanation: