Answer:
The timber companies removing all the trees from entire hillside when they are harvesting logs is a practice that could cause the following when it is time to plant in spring:
1. It could affect the quality of plant needed nutrients in the soil and beneficial microorganism population in the soil which could impact negatively the planting season.
2. Trees serves as protection of soil nutrients against wind erosions, so the soil nutrients would be affected.
The farmers tilling the soil means preparing the soil for a farming season as the following effect:
1. Helps control weed for the planting season.
2. This practice could further encourage soil erosion if not done well.
3. Tilling the soil could make leftover plants from the felling of trees mix well with the soil and also add nutrients to the soil when they decay.
Answer:
Frequency, f = 0.6 Hz
Explanation:
We have,
Number of waves passing through a point are 3
Time for which the waves are passing is 5 seconds
It is required to find the frequency of a wave. The frequency of a wave is defined as the no of waves per unit time. So,
So, the frequency of a wave is 0.6 Hz.
Answer: a. two substances present; two phases present : Heterogeneous mixture
b. two substances present; one phase present
: Homogeneous mixture
c. one substance present; one phase present
: pure substance.
d. one substance present; two phases present: Heterogeneous mixture
Explanation:
A pure substance is a substance which contains definite composition of only one type of component. Hence, it cannot be separated by physical means.
Mixture is a substance which contains two or more than two types of components and they can be separated by physical means as well.
Homogeneous mixtures: It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different. There is no physical boundary between the dispersed phase and dispersion medium.
Heterogeneous mixtures: It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different. There is a physical boundary between the dispersed phase and dispersion medium.
When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.
According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.
150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:
Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.
When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
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