In, 1937 Lawrence, in operating his cyclotron, bombarded a molybdenum-96 foil with deuterium ions (2h), producing for the first time an element not found in nature. He was initially unaware that the radioactivity produced by the "bombarded foil" was not from molybdenum but from a new, artificial element. It was his cooperation with Italian-American physicist <span>Emilio Segrè </span>that allowed the new element to be discovered. The answer is Technetium: Tc
Answer:
There are 12 nanoseconds in
.
Explanation:
We need to convert
to nanoseconds.
We know that,
![1\ s=10^9\ ns](https://tex.z-dn.net/?f=1%5C%20s%3D10%5E9%5C%20ns)
Now using unitary method to solve it such that,
![1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-8%7D%5C%20s%3D1.2%5Ctimes%2010%5E%7B-8%7D%5C%20%5Ctimes%2010%5E9%5C%5C%5C%5C%3D1.2%5Ctimes%2010%5C%5C%5C%5C%3D12%5C%20ns)
So, there are 12 nanoseconds in
.
Answer : The given compound belongs to ether and alcohol.
Explanation :
The chemical formula of the given compound is, ![C_5H_{12}O_2](https://tex.z-dn.net/?f=C_5H_%7B12%7DO_2)
First we have to calculate the degree of unsaturation.
Formula used:
Degree of unsaturation = ![\frac{2C+2+N-X-H}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2C%2B2%2BN-X-H%7D%7B2%7D)
where,
C = number of carbon
H = number of hydrogen
N = number of nitrogen
X = number of halogen
Degree of unsaturation = ![\frac{2\times 5+2+0-0-12}{2}=0](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Ctimes%205%2B2%2B0-0-12%7D%7B2%7D%3D0)
The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.
Thus, the given compound belongs to ether and alcohol.
A flow of a electric charge.
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.