'The Sedimentary rock formed from years of sediments piling on top of it and being compressed.'
1.3 x 10¹⁵ : 1.2 x 10⁹ = 1.083 x 10⁶
Answer : Option E) 50 grams.
Explanation : According to the solubility curves the compound 
to dissolve at 50 °C in 100 mL of water will need 50 grams of the compound. It is clearly indicated in the graph which is marked with red that at 50°C approximately 50.4 grams of the compound will be needed to dissolved in 100 mL of water to form a solution.
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
<h3>Further explanation</h3>
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g
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.
