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vekshin1
3 years ago
6

What is the wavelength of electromagnetic radiation with a frequency of 5.00 x 10^12 Hz? S

Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:

Infrared

Explanation:

To calculate the wavelength if the electromagnetic radiation, we must use the following formula:

c = f × λ

where c is the speed of the wave (in this case it's the speed of light, which is 3 × 10^{8} m/s) , f is the frequency and λ is the wavelength.

Using this formula, all we have to do is plug in our values and rearrange for the answer:

c = f × λ

3 × 10^{8} = 5.00 x 10^{12} Hz × λ

λ = 3 × 10^{8}/5.00 x 10^{12}

λ = 6.00 × 10^{-5} m = 6.00 ×10^{4} nm

Now that we know the wavelength we can figure out which kind of electromagnetic radiation this is.

6.00 ×10^{4} nm falls in the range of infrared radiation. We know this either from memorizing the values of the wavelengths or looking up a reference.

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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A catalyst
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A catalyst speeds up the rate of reaction so the answer is B.
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