All categories

4 years ago

Calculate [H3O+] of a 0.150 M acetic acid solution.

Physics

1 answer:

Butoxors [25]4 years ago

6 0

Answer:

[H3O+] = 0.00520 M

Explanation:

The dissociation of acetic acid in water is given by the equation;

CH3COOH + H2O<-------> H3O+ + CH3COO-

ka = [H+][C2H3O2-]/[HC2H3O2]

Ka = 1.8 x 10^-5

= x^2 / 0.150-x

x = [H3O+]= 0.00520 M

You might be interested in

A coin of mass 0.05 kg is placed at a radius r= 0.1m meter on a horizontal turntable which is turning uniformly at a rate of one

Vsevolod [243]

Answer:F=0.0882kg

Explanation:

The period it takes to make one revolution is 1.5 seconds / revolutions,

v = r * (change in angle / change in time)

the change in angle is 2pi, for one whole revolution. the time is 1.5 second per revolution, and the radius is 0.1.

v = (2 * pi * 0.1 cm * / 1.5second

v = 0.42m/s

a=v^2/r

a=0.42^2 /0.1 =1.764m/s2

F=ma

F=0.05*1.764

F=0.0882kg

4 0

3 years ago

The frequency of a microwave signal is 9.76 ghz. what is its wavelength? (c = 3.00 × 108 m/s)

dolphi86 [110]

The basic relationship between frequency of an electromagnetic wave and wavelength of the wave is
$c = \lambda f$
where $c=3 \cdot 10^8 m/s$ is the speed of light.

Manipulating the equation, we can rewrite it as
$\lambda = \frac{c}{f}$

The frequency of the wave in our problem is
$9.76 GHz = 9.76 \cdot 10^9 Hz$
so if we use the previous formula, we find the correspondant wavelength:
$\lambda = \frac{3 \cdot 10^8 m/s}{9.76 \cdot 10^9 Hz} =0.031 m$

8 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago

Based on the model, which statement best explains why point A is

Vladimir79 [104]

It’s D.Earths northern hemisphere is tilted toward the sun

6 0

3 years ago

Read 2 more answers

Calculate [H3O+] of a 0.150 M acetic acid solution.​

Calculate [H3O+] of a 0.150 M acetic acid solution.