Answer:
Molecular formula = C₄H₆As₆Cu₄O₁₆
Explanation:
Given data:
Empirical formula = C₂H₃As₃Cu₂O₈
Molar mass of compound = 1013 g/mol
Molecular formula = ?
Solution:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass of C₂H₃As₃Cu₂O₈ is 506.897 g/mol
by putting values.
n = 1013 / 506.897
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₂H₃As₃Cu₂O₈)
Molecular formula = C₄H₆As₆Cu₄O₁₆
The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.
A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.
The actual electronic configuration of antimony is as follow:
[Kr] 4d10 5s2 5p3
B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.
The actual electronic configuration of Cesium is as follow:
[Xe] 6s1
Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
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Answer:
See explanation
Explanation:
The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.
As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.
Answer:

Explanation:
Hello.
In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

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