1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
1 answer:
Answer:
Explanation:
Hello,
In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:
Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:
Finally, the mass is computed by using its molar mass (106 g/mol)
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Answer:
Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
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