The percentage of water in an unknown hydrate was determined by heating the sample and driving the water off the sample. two independent measurements gave values of 19.564% and 21.731%. its percentage should be reported as:

It should be reported as 20.648%

Step-by-step explanation:

Since we have 2 different observed values hence we shall use an average of the 2 values to report the result

Thus value is

As we can see that the least count of our observations is upto 3 decimal places hence we have to report a result upto only 3 decimal places thus we need to round off the fourth decimal place thus the digit shall be increased by 1 since we have to drop off 5 and the digit before 5 is 7 which is an odd number.

Thus the result shall be 20.648%

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632