Answer:
This question is incomplete, the complete part of the question is as follows:
This best demonstrates which type of an interaction between the plants?
A. cooperation
B. parasitism
C. commensalism
D. competition
The answer is D
Explanation:
Organisms in their natural environment interact with one another in so many ways. The ways by which this interaction occurs are; competition, predation, commensalism etc.
Competition is the interaction between two organisms in which one or both organisms are harmed due to limited resources. Competition occurs when the organisms involved occupy the same niche or utilize the same limited resources.
In this question involving corn plants and milkweed plants. They are said to grow in the same area. Over several years, the milkweed plants have taken over the field and the corn plants no longer have space to grow. In this case, there is a limited space for growth, hence, the corn plant and milkweed COMPETE.
ANSWER: A
EXPLANATION:
THE FORMULA FOR DENSITY IS MASS DIVIDED BY VOLUME.
D= M/V
So divide the mass by the volume.
For A, 122 divided by 10.5 is 11.62
For B, 132 divided by 14.2 is 9.295
For C, 115 divided by 16.1 is 7.143
For D, 126 divided by 12.7 is 9.921
A IS THE ANSWER BECAUSE ITS THE LARGEST AMOUNG THE OTHERS.
I'd say b, but i'm not 100 percent sure.<span />
Answer:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Explanation:
To answer this question we need first to understand that for organic compounds:
a. Non polar compounds have lower boiling points than polar ones of similar structure and molecular weight.
b. Boiling points increase with molecular weight. In alkane compounds if we compare isomers, the straight chain isomer will have a higher boiling point than the branched one (s) because of London dispersion intermolecular forces.
a. The introduction of hydroxyl groups increase the intermolecular forces and hence the boiling points because the electronegative oxygen, and, more importantly the presence of hydrogen bonds.
Considering the observations above, we can match the boiling points as follows:
CH₄ - 162 ⁸C
CH₃CH₃ -88.5 ⁸C
(CH₃)₂ CHCH₂CH₃ 28 ⁸C
CH₃3(CH2)₃CH₃ 36 ⁸C
CH₃OH 64.5 ⁸C
CH₃CH₂OH 78.3 ⁸C
CH₃CHOHCH₃ 82.5 ⁸C
C₅H₉OH 140 ⁸C
C₆H₅CH₂OH 205 ⁸C
HOCH₂CHOHCH₂OH 290 ⁸C
Note: There was a mistake in the symbols used for the 162 and 88.5 values which are negative and correspond to the common gases methane and ethane