Hydropower can affect fish and other marine life which is the true statement and is denoted as option C.
<h3>What is Hydropower?</h3>
This is the process in which water energy is converted into electrical energy through the use of dams and turbines.
When the dams are created, it blocks the path of marine activities and limits them to certain activities such as reproduction.
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A. : In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.
B. : In this reaction, the product is a solid(insoluble). So, this is a precipitation reaction too.
C.: In this reaction, both the products are soluble. So this is not a precipitation reaction.
D.: In this reaction, both the products are soluble. So this is not a precipitation reaction.
E. : In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.
7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.
Explanation:
Data given:
volume of phosphoric acid, Vacid =60 ml
volume of barium hydroxide, Vbase = 225 ml
molarity of barium hydroxide, Mbase = 2M
Molarity of phosphoric acid, Macid =?
the formula for titration is used as:
Macid x Vacid = Mbase x Vbase
rearranging the equation to get Macid
Macid = 
Macid =
Macid = 7.5 M
the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.
The answer is wedge to your answer
Answer:
44 g oxygen are needed.
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of ammonia = 18.2 g
Solution:
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will calculate the number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 18.2 g/ 17 g/mol
Number of moles = 1.1 mol
Now we will compare the moles of ammonia with oxygen from balance chemical equation.
NH₃ : O₂
4 : 5
1.1 : 5/4×1.1 = 1.375 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.375 mol × 32 g/mol
Mass = 44 g