Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1
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Answer:
The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.
Explanation:
Given that,
The mass of a silver liberty was measured three times and each measurement was made to five digits.
The mass values are,
The average mass of the gold coin is 12.512 g.
We know that,
Significant figures :
Significant figures is explained the nonzero, leading zero and zeros between two nonzero digits.
For example,
The digits is 0.003405
In the digit, 345 = nonzero
0.00 = leading zero
The average mass of the gold coin reported to five significant figures because average mass is 12,512. All digits are nonzero.
Nonzero digit are significant.
Hence, The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.