May somone please show me how to get the correct answer/answer range below and w
ID: 1084233 • Letter: M
Question
May somone please show me how to get the correct answer/answer range below and where the 0.04 came from?
Pretend the equation you obtained for your calibration curve is y = 1.011x + 1.1 and that you correctly plotted the midpoint of each melting range on the x-axis and their corresponding literature values on the y-axis.
You record the melting range of an unknown sample to be 96.5-98.3 oC. What is the corrected value for when melting starts? (Enter the numeric portion of your answer only. Use 1 decimal place.)
0.04 (98.66 - 98.74)
The corrected value for when melting starts is 98.7 oC.
Pretend the equation you obtained for your calibration curve is y = 1.011x + 1.1 and that you correctly plotted the midpoint of each melting range on the x-axis and their corresponding literature values on the y-axis.
You record the melting range of an unknown sample to be 96.5-98.3 oC. What is the corrected value for when melting starts? (Enter the numeric portion of your answer only. Use 1 decimal place.)
98.7
Answer range +/-0.04 (98.66 - 98.74)
Response Feedback:The corrected value for when melting starts is 98.7 oC.
Explanation / Answer
Given equation of line is y = 1.011x + 1.1
Where y corresponds to the literature values of melting point
x corresponds to mid-point of each melting range
i.e. The starting value: y = 96.5*1.011 + 1.1
i.e. y = 97.5615 + 1.1
i.e. y = 98.66
If you round the above value to 1 decimal place, you will get y = 98.7
Here, the error = 98.7 - 98.66 = +0.04
For suppose, y = 98.74, then also, if you round it to 1 decimal place, you will get y = 98.7
Here, the error = 98.7 - 98.74 = -0.04
Therefore, the answer range could be +/- 0.04