true
We describe the luminescence spectral properties of CdS nanoparticles with multiphoton excitation. Three types of CdS nanoparticles were examined which were a CdS/dendrimer composite which displays high anisotropy, Cd2+-enriched nanoparticles which display two emission maxima, and polyphosphate-stabilized nanoparticles which display long wavelength emission. Illumination with long wavelengths near 700−790 nm resulted in two-photon excitation. Essentially the same emission spectra and intensity decays were observed with one-photon and two-photon excitation. Comparison with fluorescein indicates the NPs display large two-photon cross sections near 100 GM. The CdS/dendrimer and Cd2+-enriched CdS nanoparticles displayed large anisotropy values with two-photon excitation, substantially larger than with one-photon excitation. It appears that semiconductor nanoparticles are comparable to organic fluorophores which display the same spectral properties with one-photon and two-photon excitation.
The first fact you should know to solve a stoichiometric problem like this one is that, there are 6.02 x 10^23 particles (atoms or molecules) in 1 mole of a substance.
Now, to know the equivalence factor, it is important to know a technique which cancels like units when they are placed diagonally. Hence, to find the moles Mg from atoms, the equivalence factor must have a numerator with units of moles and a denominator with units of atoms. In this case, atoms would be cancelled, leaving the moles.
<u>Answer:</u> The final volume reading on the base is 72.35 mL
<u>Explanation:</u>
To calculate the volume of base, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![H_2SO_4](https://tex.z-dn.net/?f=H_2SO_4)
are the n-factor, molarity and volume of base which is ![Ba(OH)_2](https://tex.z-dn.net/?f=Ba%28OH%29_2)
We are given:
![n_1=2\\M_1=0.472M\\V_1=[41.90-34.42]=7.48mL\\n_2=2\\M_2=0.388M\\V_2=?mL](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D0.472M%5C%5CV_1%3D%5B41.90-34.42%5D%3D7.48mL%5C%5Cn_2%3D2%5C%5CM_2%3D0.388M%5C%5CV_2%3D%3FmL)
Putting values in above equation, we get:
![2\times 0.472\times 7.48=2\times 0.388\times V_2\\\\V_2=\frac{2\times 0.472\times 7.48}{2\times 0.388}=9.10mL](https://tex.z-dn.net/?f=2%5Ctimes%200.472%5Ctimes%207.48%3D2%5Ctimes%200.388%5Ctimes%20V_2%5C%5C%5C%5CV_2%3D%5Cfrac%7B2%5Ctimes%200.472%5Ctimes%207.48%7D%7B2%5Ctimes%200.388%7D%3D9.10mL)
Final volume of barium hydroxide solution = Initial volume + Volume needed to neutralize = [63.25 + 9.10] = 72.35 mL
Hence, the final volume reading on the base is 72.35 mL
1: viewing any chemical reaction in a laboratory
2: dangerous to look at when it burns & used in photography, fireworks, and flares
3: the product