If you dilute 250 mL of 6.4 M lithium acetate solution to a volume of 750 mL, what will the concentration of this solution be?

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

What mass of silver chloride can be produced from 1.82 l of a 0.176 m solution of silver nitrate?

Fofino [41]

Not sure as don't know ratios, I think it could be 45.93g but don't take my word for it, it could be wrong.

The equations you need are moles = concentration x volume

and mass = moles x formula mass

5 0

3 years ago

Dew forming on grass is an example of condensation true or fales and explain

exis [7]

True becuase dew is coming out of the air which if you look at a glass of water it has condensation on it becuase it is hot

7 0

3 years ago

Read 2 more answers

How many moles are in 442.8 g of strontium carbonate (SrCO3)?

GuDViN [60]

Answer:

3 moles

Explanation:

SrCO3

Mass = 442.8g

Molar mass = (87.6 + 12 * [16*3]) = 147.6g/mol

Number of moles = mass / molar mass

Number of moles = 442.8 / 147.6

Number of moles = 3

7 0

3 years ago

Part A Which acid in each of the following pairs has the stronger conjugate base? Match the words in the left column to the appr

vichka [17]

Answer:

HF

H₂S

H₂CO₃

NH₄⁺

Explanation:

Which acid in each of the following pairs has the stronger conjugate base?

According to Bronsted-Lowry acid-base theory, the weaker an acid, the stronger its conjugate acid. Especially for weak acids, pKa gives information about the strength of such acid. The higher the pKa, the weaker the acid.

Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.

Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39

Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.

Of the acids HF or NH₄⁺, the one with the stronger conjugate base is NH₄⁺ because it is a weaker acid. pKa (HF) = 3.17 < pKa (NH₄⁺) = 9.25

6 0

3 years ago