<u>Answer:</u> The half life of the reaction is 593.8 seconds
<u>Explanation:</u>
We are given:
Rate constant = ![0.0016mol/L.s](https://tex.z-dn.net/?f=0.0016mol%2FL.s)
The formula for determining the unit of 'k' is:
![\text{Unit}=\frac{(Concentration)^{1-n}}{Time}](https://tex.z-dn.net/?f=%5Ctext%7BUnit%7D%3D%5Cfrac%7B%28Concentration%29%5E%7B1-n%7D%7D%7BTime%7D)
where, n = order of reaction
The unit of concentration is, M or mole/L
The unit of time is, second or 's'
Evaluating the value of 'n' from above equation:
![mol.L^{-1}s^{-1}=\frac{(mol/L)^{1-n}}{s}\\\\n=0](https://tex.z-dn.net/?f=mol.L%5E%7B-1%7Ds%5E%7B-1%7D%3D%5Cfrac%7B%28mol%2FL%29%5E%7B1-n%7D%7D%7Bs%7D%5C%5C%5C%5Cn%3D0)
The reaction is zero order reaction.
The equation used to calculate half life for zero order kinetics:
![t_{1/2}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)
where,
k = Rate constant = ![0.0016mol/L.s](https://tex.z-dn.net/?f=0.0016mol%2FL.s)
= initial concentration = 1.90 mol/L
Putting values in above equation, we get:
![t_{1/2}=\frac{1.90mol/L}{2\times 0.0016mol/L.s}=593.8s](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1.90mol%2FL%7D%7B2%5Ctimes%200.0016mol%2FL.s%7D%3D593.8s)
Hence, the half life of the reaction is 593.8 seconds
Answer:
![{256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1](https://tex.z-dn.net/?f=%20%7B256x%7D%5E%7B4%7D%2B%7B256x%7D%5E%7B3%7D%20%2B%20%7B96x%7D%5E%7B2%7D%20%2B%2016x%20%2B%201%20)
Explanation:
![{(1 + 4x)}^{4} = (1 + 4x)(1 + 4x)(1 + 4x)(1 + 4x)](https://tex.z-dn.net/?f=%20%7B%281%20%2B%204x%29%7D%5E%7B4%7D%20%20%3D%20%281%20%2B%204x%29%281%20%2B%204x%29%281%20%2B%204x%29%281%20%2B%204x%29)
The coefficient of the kth term (ordering in increasing order for the exponent of x) is just the number of ways we have to choose k factors from that expression, so if we let k be the exponent of x, and n be the total number of terms, the coefficient of x^k is ![\binom{n}{k} = \frac{n!}{k!(n-k)!}](https://tex.z-dn.net/?f=%20%5Cbinom%7Bn%7D%7Bk%7D%20%20%3D%20%20%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D%20)
Which, of course, we have to multiply for the product of the two terms.
For example, the coefficient of the third grade term in
is
So we have ![{(1 + 4x)}^{4} = \binom{4}{4} \cdot {(4x)}^{4}+\binom{4}{3} \cdot {(4x)}^{3} + \binom{4}{2} \cdot {(4x)}^{2} + \binom{4}{1} \cdot 4x + \binom{4}{0} \cdot 1](https://tex.z-dn.net/?f=%20%7B%281%20%2B%204x%29%7D%5E%7B4%7D%20%3D%20%5Cbinom%7B4%7D%7B4%7D%20%5Ccdot%20%7B%284x%29%7D%5E%7B4%7D%2B%5Cbinom%7B4%7D%7B3%7D%20%5Ccdot%20%7B%284x%29%7D%5E%7B3%7D%20%2B%20%5Cbinom%7B4%7D%7B2%7D%20%5Ccdot%20%7B%284x%29%7D%5E%7B2%7D%20%2B%20%5Cbinom%7B4%7D%7B1%7D%20%5Ccdot%204x%20%2B%20%5Cbinom%7B4%7D%7B0%7D%20%5Ccdot%201%20%20)
Which is equal to ![1 \cdot {(4x)}^{4}+4 \cdot {(4x)}^{3} + 6 \cdot {(4x)}^{2} + 4 \cdot 4x + 1 \cdot 1 = {256x}^{4}+{256x}^{3} + {96x}^{2} + 16x + 1](https://tex.z-dn.net/?f=%201%20%5Ccdot%20%7B%284x%29%7D%5E%7B4%7D%2B4%20%5Ccdot%20%7B%284x%29%7D%5E%7B3%7D%20%2B%206%20%5Ccdot%20%7B%284x%29%7D%5E%7B2%7D%20%2B%204%20%5Ccdot%204x%20%2B%201%20%5Ccdot%201%20%20%3D%20%7B256x%7D%5E%7B4%7D%2B%7B256x%7D%5E%7B3%7D%20%2B%20%7B96x%7D%5E%7B2%7D%20%2B%2016x%20%2B%201%20)
Hope this helps :)
Answer:
12 mmilligrams of Po-218 was the mass of the original starting material
Explanation:
The half-life of a radioactive material is the time taken for half the amount ofnthe original material present in a radioactive material to decay or disintegrate.
After each half-life, half the original material present at the start remains.
For the radioactive polonium-218 having a half-life of 3.04 minutes, it means that if 1 g is the starting material, after 3.04 minutes, 1/2 g will be remaining; after, 6.08 minutes 1/2 of 1/2 which is 1/4 of the starting material will be remaining; and after 9.12 minutes, 1/2 of 1/4 = 1/8 g will be remaining.
From the question, number of half-lives undergone after 9.12 minutes = 9.12/3.04 = 3 half-lives.
After 3 half-lives, 1/8 of the original material is remaining.
1/8 = 1.50 mg
The original mass of the sample at the start = 1.50 mg × 8 = 12 mg
Therefore, 12 milligrams of Po-218 was the mass of the original starting material.
C. Carbon tetra–bromide is a non–polar compound, for the four Bromines balance out the dipole moment from the tetrahedral bonds they form on Carbon.
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