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2 years ago

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

Physics

2 answers:

tia_tia [17]2 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]2 years ago

3 0

<span>A.) The hydronium ion concentration in the solution.</span>

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3 years ago

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2 years ago

Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l

ziro4ka [17]

Answer:

The pressure is 6570 lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

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Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

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$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570 lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$

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3 years ago