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gulaghasi [49]
3 years ago
12

(scientific investigations) help please.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

I agree with what you've chosen for question two and three.

I would choose C for the fourth question and argue that you shall also go with C for question five.

One of the main ideas about repetition is to have multiple <em>Trials</em>- or independent experiments-<em> </em>for a single independent variable using identical experimental setup. The table presented in option C is the only one among the four choices that allots spaces for measurements from multiple independent experiments.

Hypotheses are predictions of the outcome of a particular experiment. A hypothesis is <em>correct</em> only if it accurately describes the outcome of an experiment. The <em>hypothesis</em>, not the data, would have to be revised if the two doesn't fit. The researcher could keep developing and testing new hypotheses until arriving at a correct one.

Additionally, I would prefer option B over option D in the first question- responses to that question can depends on what your teacher says about the validity of different sources, given that <em>neither </em><em>B</em><em> nor </em><em>D</em><em> </em> are reliable as sources that one would like to cite in a paper.

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What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

pH=7.00

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