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jasenka [17]
3 years ago
13

Which of the following is a disadvantage of using hydropower? (2 points)

Chemistry
2 answers:
frutty [35]3 years ago
7 0
D.) Land can be flooded, displacing people and wildlife.
anygoal [31]3 years ago
4 0

Answer : The disadvantage of using hydropower is Land can be flooded, displacing people and wildlife

Explanation :

Hydropower is a renewable source of energy. In hydropower, electricity is generated using the mechanical energy of moving water.

Since the kinetic energy is necessary to move the turbines, it is essential that the hydropower plant is present near a water source which has moving water.

It is the most efficient way of harnessing the energy from naturally available energy sources. It does not produce any waste. It does not pollute water or air.

Since hydropower needs kinetic energy of moving water, there is always a possibility of nearby land getting flooded. This would affect the people and wildlife of nearby areas.

Therefore the disadvantage of using hydropower is that land can be flooded, displacing people and wildlife


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Explain how the study of taxonomy helps other scientists.
katrin [286]
The classification of living things makes it easier for scientists to answer many important questions.
Examples:
-How many known species are there?
-What are the defining characteristics of each species?
-What are the relationships between these species?
4 0
3 years ago
Surface tension of solid,liquid and gas​
Arisa [49]

Explanation:

<h3>Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects, usually denser than water, to float and slide on a water surface.</h3>

4 0
3 years ago
Which of the following gives the molarity of a 17.0% by mass solution of sodium acetate, CH 3COONa (molar mass = 82.0 g/mol) in
weeeeeb [17]

Answer:

The molarity of this solution is 2.26 M (option D)

Explanation:

Step 1: Data given

Mass % = 17%

Molar mass of CH3COONa = 82.0 g/mol

Density of the solution = 1.09 g/mL

Step 2:

Assume the mass of the solution is 1.00 gram

⇒ 17.0 % CH3COONa = 0.17 grams

⇒ 83.0 % H2O = 0.83 grams

Step 3:

Density = 1.09 g/mL

Volume of the solution = total mass / density of solution

Volume of solution : 1.00 grams / 1.09 g/mL

Volume of the solution = 0.917 mL = 0.000917 L

Step 4: Calculate  number of moles of CH3COONa

Number of moles = Mass / molar mass

Number of moles CH3COONa = 0.170 grams / 82.0 g/mol

Number of moles = CH3COONa = 0.00207 moles

Step 5: Calculate molarity

Molarity = moles / volume

Molarity solution =   0.00207 / 0.000917 L

Molarity solution =  2.26 mol/L = 2.26 M

The molarity of this solution is 2.26 M (option D)

6 0
3 years ago
How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat
stira [4]

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

8 0
3 years ago
What is the new solution concentration when 150. mL of water is added to 200. mL of a 3.55 M HBr solutions.
Agata [3.3K]

Answer:

The new concentration is 2.03M

Explanation:

Step 1: Data given

A 200 mL 3.55 M HBr is diluted with 150 mL

Step 2: The dilution

In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume

In this case, the volume of the stock solution is 200 mL

Adding  150 mL  of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL

The dilution factor wll be 350/200 = 1.75

This makes the diluted concentration:

3.55/1.75 = 2.03M

The new concentration is 2.03M

3 0
3 years ago
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