Some of the NH₄+ will combine with the OH- and shift the equilibrium backwards and from NH₄OH to balance the change produced by addition of NH₄+ ions.
Answer:
Wave energy is the transport and capture of energy by ocean surface waves. The energy captured is then used for all different kinds of useful work, including electricity generation, water desalination, and pumping of water.
Explanation:
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Answer:</h3>
2.125 g
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Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
Learn more about equilibrium of forces here:
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