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3 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]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:

I think it's 10 minutes per km

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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Nookie1986 [14]

<h2>Answers:</h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ are constants:

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

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4 years ago

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Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
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8a.The mass of a girl is 40 kg. Calculate her weight. (g = 9.8 m/s)

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Explanation:

Here,

Given,

Mass(m)=40 kg

Gram=9.8m/s

Now,

Weight=m x g

or, weight= 40x9.8

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QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

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