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2 years ago

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Chemistry

1 answer:

Katarina [22]2 years ago

5 0

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:

brainly.com/question/12980489

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2 years ago

Which bonds form in the reaction shown in the diagram?

lesya692 [45]

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2 years ago

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2 years ago

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Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$

[HB] is the concentration of base.

$8 * 10^{-5} =\frac{x^{2} }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1$

$x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}$

Then

$pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54$

So the pH of the solution is 2.54.

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3 years ago