Answer:
Yes, Pb3(PO4)2.
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.
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Answer:
Option C = electron
Explanation:
Electrons are responsible for the production of colored light.
Electron:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol= e-
Mass= 9.10938356×10⁻³¹ Kg
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
How electrons produce the colored light:
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.
Other process may involve,
Fluorescence:
In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.
The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.
Phosphorescence:
In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.
Yes because look in the book dh
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
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