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Alex777 [14]
3 years ago
11

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

m/s in 3 seconds, what is the acceleration of the car?
2) While cleaning you lift an 87.3 N box up to a shelf that is 2.04 m above the ground. How much work was done on the box?

3) A child runs up the stairs and does 1250 J of work. If the child generates 267 W of power how long did it take for the child to run up the stairs?

4) A 8.642 kg rabbit is running across your back yard. If that rabbit has 125.6 J of Kinetic Energy, how fast is it running?

5) When using a simple machine you do 120 J of work on the machine. You notice that the machine gives you an output of 93 J. What is the efficiency of this machine?

6) A 212 kg bumper car is traveling 8.00 m/s when it rear ends a second 196 kg bumper car traveling 6.75 m/s. If the cars get stuck together, how fast are they traveling after the collision?

7) A wave is traveling through a string. If the wave has a wavelength of .23 m and a frequency of 12 Hz, what is the speed of the wave?

8) Find the wavelength of blue light if its frequency is 6.2 × 1014 Hz. (Hint: Remember the speed of light)

9) The speed of sound in air is about 343 m/s. If you scream and a friend hears you 0.287 s later, how far away from you is your friend?
Chemistry
1 answer:
WITCHER [35]3 years ago
3 0

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

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Answer:

See explanations

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversations:

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation Na + H₂O => NaOH + H₂. ‘Reading the equation’ there is 1 mole of Na, 1 mole of water, 1 mole of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation Na + H₂O => NaOH + H₂, one could apply a 2 before the Na but all the following formulas would need be increased by a factor of 2. If one applies ½ to the Na then all the following formulas would need be cut in half also and the reaction stoichiometry would still be valid. The fact that the equation is written with coefficients of 1 is that it is in the smallest whole number ratio of coefficients. This then implies the reaction formula is in ‘standard form’. This also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.  

For your 1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

                         Na                      +            H₂O       => NaOH    +         H₂,

Given:      1.76 x 10²⁴ atoms                excess             ---------          ? grams

Convert atoms Na to moles = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

=2.922moles Na produces=>2.922moles H₂(because moles Na=moles H₂).

Convert moles to grams =>2.922moles H₂  x  2.000 grams H₂/mole H₂

=5.8443 grams H₂  

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again write equation and assign values to each formula unit in the equation.

                         Na                      +            H₂O        =>           NaOH    +    H₂,

Given:            3.5moles                       excess                      ? grams       ----

Since coefficients of balanced std equation are equal then moles Na equals moles of NaOH, that is, 3.5 moles Na produces => 3.5 moles NaOH

Convert moles NaOH to grams => 3.5 moles NaOH x 40 g NaOH/mole NaOH =  140 grams NaOH    

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

                         Na          +            H₂O                              =>       NaOH    +    H₂

Given:           ? atoms          2.75 x 10²⁵ molecules H₂O    =>     NaOH  + H₂  

Convert to moles =>  2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O = 45.658 moles H₂O =>  45.658 moles Na (equal coefficients)

Convert moles Na to atoms Na  =>   45.658 moles Na x 6.023 x 10²³atoms Na/mole Na = 2.75 x 10²⁵ atoms Na.

Note => Problem 3 could have been solved by inspection b/c coefficients are equal, however, always go through a process that you can justify and defend even if it does take longer. Never assume anything. Depend on what you know, not what you 'think' you know.  

Master the mole concept and you master a lot of chemistry! Good luck.

                             

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Explanation:

Initial concentration of CoCl_2 = 0.056 M

Initial concentration of Cl^- = 4.60 M

The given balanced equilibrium reaction is,

               COCl_2+2Cl^-\rightleftharpoons [CoCl_4]^{2-}+6H_2O

Initial conc.          0.056 M      4.60 M        0 M       0 M

At eqm. conc.     (0.056-x) M   (4.60-2x) M   (x) M    (6x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CoCl_4]^{2-}\times [H_2O]^6}{[CoCl_2]^2\times [Cl^-]^2}

Given : equilibrium concentration of [CoCl_4]^{2-} =x =  0.031 M

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Here, the Pressure is decreasing from 537 torr to 255 torr. So, the volume must increase and calculated by using the equation:

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Let's plug in the values in the equation:

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V_2=(\frac{537 torr*5.30 L}{255 torr})

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