Enrichment of uranium-235 is done. Hence, option A is correct.
<h3>What is the meaning of containment?</h3>
The act, process, or means of keeping something within limits the containment of health costs.
Enrichment is a process of increasing the proportion of fissile isotope found in uranium ore (represented by the symbol 'U') to make it usable as nuclear fuel or the compressed, explosive core of nuclear weapons.
Hence, option A is correct.
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Answer:The electron configuration of an atom shows the number of electrons in each sublevel in each energy level of the ground-state atom. To determine the electron configuration of a particular atom, start at the nucleus and add electrons one by one until the number of electrons equals the number of protons in the nucleus. Each added electron is assigned to the lowest-energy sublevel available. The first sublevel filled will be the 1s sublevel, then the 2s sublevel, the 2p sublevel, the 3s, 3p, 4s, 3d, and so on. This order is difficult to remember and often hard to determine from energy-level diagrams such as Figure 5.8
A more convenient way to remember the order is to use Figure 5.9. The principal energy levels are listed in columns, starting at the left with the 1s level. To use this figure, read along the diagonal lines in the direction of the arrow. The order is summarized under the diagram
Answer:
Number of moles = 0.042 mol
Explanation:
Given data:
Number of moles = ?
Mass of calcium carbonate = ?
Solution:
Formula:
Number of moles = mass/ molar mass
now we will calculate the molar mass of calcium carbonate.
atomic mass of Ca = 40 amu
atomic mass of C = 12 amu
atomic mass of O = 16 amu
CaCO₃ = 40 + 12+ 3×16
CaCO₃ = 40 + 12+48
CaCO₃ = 100 g/mol
Now we will calculate the number of moles.
Number of moles = 4.15 g / 100 g/mol
Number of moles = 0.042 mol
The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂