Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate both moles and grams of lanthanum by using the Avogadro's number as a relationship of atoms to moles and its atomic mass as a relationship to moles to grams to obtain the following:
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These are buffer systems (solutions of substances) with a steady concentration of hydrogenium ions. At addition to such solution of a small amount of acid or alkali does not change рН.
For example:
CH₃COOH + CH₃COONa acetate buffer
KH₂PO₄ + K₂HPO₄ potassium phosphate buffer
All of the acid molecules in beaker 1 dissociate fully and exist as and ions. As a result, beaker 1 represents a strong acid solution. The majority of the molecules in beaker 2 are undissociated.
Answer:
5.95 L
Explanation:
V1/V2 = T1/T2
4.82 L/V2 = 243.1K/300.1K
V2 = 4.82*300.1/243.1 L = 5.95 L
