What are three things that can generate electrical energy?

attashe74 [19]

A quart is roughly 2lbs.

8 0

3 years ago

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Rom4ik [11]

M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

M(HCl) = 0.35
V(HCl) = 45mL
M(NaOH)= 0.35

now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

0.35 * 45 = 0.35 * V(NaOH)

V(NaOH) = 45 mL

8 0

3 years ago

Read 2 more answers

What is the concentration of H+ in 0.0025 M HClO4? What is the pH of the solution? What is the OH− concentration in the solution

alexandr402 [8]

Answer:

A. The concentration of H+ is 0.0025 M

B. The pH is 2.6

C. The concentration of OH- is 3.98x10^-12 M

Explanation:

We'll begin by writing the balanced

dissociation equation of HClO4. This is illustrated below:

HClO4 —> H+ + ClO4-

A. Determination of the concentration of H+ in 0.0025 M HClO4. This is illustrated below:

From the balanced equation above,

1 mole of HClO4 produced 1 mole of H+.

Therefore, 0.0025 M of HClO4 will also produce 0.0025 M of H+.

The concentration of H+ is 0.0025 M

B. Determination of the pH.

The pH of the solution can be obtained as follow:

The concentration of H+, [H+]

= 0.0025 M

pH =?

pH = - log [H+]

pH = - log 0.0025

pH = 2.6

C. Determination of the concentration of OH-

To obtain the concentration of OH-, we must first calculate the pOH of the solution. This is illustrated below:

pH + pOH = 14

pH = 2.6

pOH =?

pH + pOH = 14

2.6 + pOH = 14

Collect like terms

pOH = 14 - 2.6

pOH = 11.4

Now, we can calculate the concentration of the OH- as follow:

pOH = - Log [OH-]

pOH = 11.4

11.4 = - Log [OH-]

- 11.4 = log [OH-]

[OH-] = anti log (- 11.4)

[OH-] = 3.98x10^-12 M

8 0

3 years ago

1) A car is traveling down the interstate at 37.1 m/s. The driver sees a cop and quickly slows down. If the driver slows to 29.8

WITCHER [35]

1)

The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity of the car to change from u to v

In this problem, for this car we have:

u = 37.1 m/s

v = 29.8 m/s

t = 3 s

So, the acceleration is:

$a=\frac{29.8-37.1}{3}=-2.43 m/s^2$

2)

The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:

$W=Fd$

where

F is the force applied

d is the displacement of the box

Here we have:

F = 87.3 N is the force applied

d = 2.04 m is the displacement of the box

So, the work done to lift the box is:

$W=(87.3)(2.04)=178.1 J$

3)

The power is the rate of work done per unit time. It is calculated as:

$P=\frac{W}{t}$

where

W is the work done

t is the time taken to do the work

For the child in this problem, we have:

W = 1250 J is the work done by the child running up the stairs

P = 267 W is the power used

Therefore, re-arranging the equation, we find the time taken:

$t=\frac{W}{P}=\frac{1250}{267}=4.68 s$

4)

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the rabbit in this problem, we have:

m = 8.642 kg is the mass of the rabbit

KE = 125.6 is its kinetic energy

Solving the formula for v, we find the speed of the rabbit:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s$

5)

The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by

$\eta = \frac{E_{out}}{W_{in}}\cdot 100$

where

$E_{out}$ is the energy in output

$W_{in}$ is the work in input

For the machine in this problem,

$W_{in}=120 J$ is the work in input

$E_{out}=93 J$ is the energy in output

Therefore, the efficiency of this machine is:

$\eta=\frac{93}{120}\cdot 100=77.5\%$

6)

During a collision, the total momentum of the system is always conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 =(m_1+m_2)v$

where

$m_1=212 kg$ is the mass of the first car

$u_1=8.00 m/s$ is the initial velocity of the first car

$m_2=196 kg$ is the mass of the 2nd car

$u_2=6.75 m/s$ is the initial velocity of the 2nd car

$v$ is the final velocity of the two cars stuck together (after the collision, they move together)

Solving the equation for v, we find:

$v=\frac{m_1 u_1 +m_2 u_2}{m_1 +m_2}=\frac{(212)(8.00)+(196)(6.75)}{212+196}=7.40 m/s$

7)

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is the frequency of the wave

$\lambda$ is the wavelength

For the wave in the string in this problem we have:

$\lambda=0.23 m$ (wavelength)

f = 12 Hz (frequency)

So, the speed of the wave is:

$v=(12)(0.23)=2.76 m/s$

8)

The relationship between frequency and wavelength for an electromagnetic wave is given by

$c=f\lambda$

where:

c is the speed of light in a vacuum

f is the frequency of the wave

$\lambda$ is the wavelength of the wave

For the blue light in this problem, we have

$f=6.2\cdot 10^{14}Hz$ (frequency)

while the speed of light is

$c=3.0\cdot 10^8 m/s$

So, the wavelength of blue light is:

$\lambda=\frac{c}{f}=\frac{3.0\cdot 10^8}{6.2\cdot 10^{14}}=4.8\cdot 10^{-7} m$

9)

The sound wave in this problem travels with uniform motion (=constant velocity), therefore we can use the following equation:

$d=vt$

where

d is the distance covered by the wave

v is the speed of the wave

t is the time elapsed

In this problem:

v = 343 m/s is the speed of the sound wave

t = 0.287 s is the time elapsed

So, the distance covered by the wave is

$d=(343)(0.287)=98.4 m$

3 0

3 years ago

Which is a nervous system test?

Elza [17]

Answer:

Sorry, is there supposed to be an attachment, If so kindly republish your Question.

3 0

3 years ago