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Answer:
C
Explanation:
You mix different thing together to make a new thing.
Answer:
most likely that (2) the replicated experiment was performed incorrectly.
Why, u ask? u dare question me:
1- The initial experiment invalidness cannot be proven.
2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>
3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.
4- Already knowing the data and errors would increase the precision of the replicated experiment.
5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.
happy learning!
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.