Answer:
3 years
Explanation:
Given data:
Initial amount of sample = 160 Kg
Amount left after 12 years = 10 Kg
Half life = ?
Solution:
at time zero = 160 Kg
1st half life = 160/2 = 80 kg
2nd half life = 80/2 = 40 kg
3rd half life = 40 / 2 = 20 kg
4th half life = 20 / 2 = 10 kg
Half life:
HL = elapsed time / half life
12 years / 4 = 3 years
Answer:
Explanation:
In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.
We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.
The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.
To ascertain the oxidation state, we have to comply with some rules:
- The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
- The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.
For example, let us find the oxidation state of Cr in Cr₂O₇²⁻
This would be: 2x + 7(-2) = -2
x = +6
We see that the oxidation number of Cr, a transition metal in the given ion is +6.
<u>Answer:</u> The energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)
<u>Explanation:</u>
To calculate the energy of one photon, we use Planck's equation, which is:
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
= wavelength of light =
(Conversion factor:
)
h = Planck's constant = ![6.625\times 10^{-34}J.s](https://tex.z-dn.net/?f=6.625%5Ctimes%2010%5E%7B-34%7DJ.s)
c = speed of light = ![3\times 10^8m/s](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E8m%2Fs)
Putting values in above equation, we get:
![E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{5.25\times 10^{-7}mm}\\\\E=3.79\times 10^{-19}J](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B6.625%5Ctimes%2010%5E%7B-34%7DJ.s%5Ctimes%203%5Ctimes%2010%5E8m%2Fs%7D%7B5.25%5Ctimes%2010%5E%7B-7%7Dmm%7D%5C%5C%5C%5CE%3D3.79%5Ctimes%2010%5E%7B-19%7DJ)
Hence, the energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)
(B. 3) 172 All nonzero digits are significant.
(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.
(A. 4) 3427 All nonzero digits are significant.
(B. 3) 0.0000455 Leading zeroes are not significant.
(B. 3) 0.00456 Leading zeroes are not significant.
(C. 5) 2205.2 Zeroes between nonzero digits are significant.
(C. 5) 107.20 Trailing zeroes after the decimal point are significant.
(B. 3) 0.0473 Leading zeroes are not significant.