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However, I can tell you how to solve this.
Answer:
<span>As the wavelength gets shorter (closer together), the frequency of the wave increases.
Explanation:
The relation between frequency and wavelength can be described by the help of velocity as follows:
velocity = frequency * wavelength
This means that:
frequency = velocity / wavelength
Noting this equation, we will find that:
The frequency and the wavelength are inversely proportional to each other. This means that as the frequency increases, the wavelength decreases and vice versa.
Now, examining the choices given, we can find that the only statement showing the inverse relation between frequency and wavelength is:
</span><span>As the wavelength gets shorter (closer together), the frequency of the wave increases.
Hope this helps :)
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It would be in the fourth shell.
Answer:
This atomic model has changed over time. Scientists used the model to make predictions. Sometimes the results of their experiments were a surprise and they did not fit with the existing model. Scientists changed the model so that it could explain the new evidence.
Explanation:
1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.
2. atomic mass of
Cu = 63.546
S = 32.065
O = 15.9994
3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.
4. 63.546 g + 32.065 g + ( 4 x <span>15.9994) = 159.609 g
5. this value </span><span>159.609 g is the mass in grams of one mol of CuSO4
6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4
7 then you have the multiply the value of one mol by the number of moles that the problem is asking you
8. </span><span>159.609 g x 3.65 = 582.571 g
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9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"