The answer would be the sound waves.
Answer : The volume of a sample of 4.00 mol of copper is ![28.5cm^3](https://tex.z-dn.net/?f=28.5cm%5E3)
Explanation :
First we have to calculate the mass of copper.
![\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20copper%7D%3D%5Ctext%7B%20Moles%20of%20copper%7D%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20copper%7D)
![\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20copper%7D%3D%284.00moles%29%5Ctimes%20%2863.5g%2Fmole%29%3D254g)
Now we have to calculate the volume of copper.
Formula used :
![Density=\frac{Mass}{Volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7BMass%7D%7BVolume%7D)
Now put all the given values in this formula, we get:
![8.92\times 10^3kg/m^3=\frac{254g}{Volume}](https://tex.z-dn.net/?f=8.92%5Ctimes%2010%5E3kg%2Fm%5E3%3D%5Cfrac%7B254g%7D%7BVolume%7D)
![Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3](https://tex.z-dn.net/?f=Volume%3D%5Cfrac%7B254g%7D%7B8.92%5Ctimes%2010%5E3kg%2Fm%5E3%7D%3D2.85%5Ctimes%2010%5E%7B-2%7DL%3D2.85%5Ctimes%2010%5E%7B-2%7D%5Ctimes%2010%5E3cm%5E3%3D28.5cm%5E3)
Conversion used :
![1kg/m^3=1g/L\\\\1L=10^3cm^3](https://tex.z-dn.net/?f=1kg%2Fm%5E3%3D1g%2FL%5C%5C%5C%5C1L%3D10%5E3cm%5E3)
Therefore, the volume of a sample of 4.00 mol of copper is ![28.5cm^3](https://tex.z-dn.net/?f=28.5cm%5E3)
Answer:
(a) T = 10 s
(b) f = 0.1 Hz
(c) λ = 32 m
(d) v = 3.2 m/s
(e) Insufficient data
Explanation:
(a)
Time period is defined as the time interval required for one wave to pass. Therefore, the time period can be given as:
T = Period = Time Taken/No. of Waves
T = 50 s/5
<u>T = 10 s</u>
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(b)
Frequency is the reciprocal of time period:
f = frequency = 1/T
f = 1/10 s
<u>f = 0.1 Hz</u>
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(c)
Wavelength is the distance between two consecutive crests or troughs:
<u>λ = Wavelength = 32 m</u>
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(d)
Speed of wave is given by the following formula:
Speed = v = fλ
v = (0.1 Hz)(32 m)
v = 3.2 m/s
(e)
Amplitude cannot be found with given data.
Your answer would be B. It forms hydrogen ions (H+)
Explanation:
m = kg. v=m/s. g=m/s^2. h= m
>>1/2mv^2=mgh
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .
<u>>>kg m^2/s^2=kg m^2/s^2</u><u> </u>
<u>></u><u>></u><u>J</u>= J