Answer:
Explanation:
What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.
Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
Answer: Option (3) is the correct answer.
Explanation:
Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.
Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.
Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.
Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.
Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.
Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.
Would the answer be d wouldn't it
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