Astudent took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and found it read 186.5 g. what was the student’s percent error?

The original or accepted value for the percent by mass of water in a hydrate = 36%

Percen by mass of water in the hydrate determined by the student

in the laboratory = 37.8%

So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%

            = 1.8%

So the percentage of error made by the student = (1.8/36) * 100 percent

                                                                            = (18/360) * 100 percent

                                                                            = ( 1/20) * 100 percent

                                                                             = 5 percent

So the student makes an error of 5%. Option "1" is the correct option.

Explanation:

the answer is b, have negligible mass

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this