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Maksim231197 [3]
3 years ago
6

Which is the most closely related to the color of an acid-base indicator when dipped into solution?

Physics
2 answers:
tia_tia [17]3 years ago
7 0

Answer : The correct option is, (A) The hydronium ion concentration in the solution.

Explanation :

Acid-base indicator : It is used in the titration. It is a substance which changes the color with the pH.

Acid-base indicator are the weak acid-base indicators which when dissolved in the water, it dissociate slightly and form the ions.

When the indicator dissolved in the water then it dissociate into H^+ ion and then the color change of the pH indicator.

Let us consider an indicator which is a weak acid, with the formula HIn.

At equilibrium, the chemical equation of acid indicator will be,

HIn \rightleftharpoons H^++In^-

Hence, the correct option is, (A) The hydronium ion concentration in the solution.

Svetlanka [38]3 years ago
3 0
<span>A.) The hydronium ion concentration in the solution.</span>
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Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

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\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

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\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

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