In 52.6 mins the 10 mg of iodine will be divided by its half-life which is 2 it will become 5 mg (in 52.6 minutes ) and same procedure will be followed until it reach 99% out of the patient's system.
The answer is 368 minutes will it takes the iodine to be absorbed in one's system.
Answer:
In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula.
Explanation:
In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula. The core remains as a white dwarf and finally become a black dwarf as it cools down. A low mass star consumes its core hydrogen and turns it into helium over its lifetime.
Answer:
Solution. Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt.
Explanation:
Answer:
weighing balance/analytical balance
Graduated cylinder/buret
Explanation:
The mass of the evaporating basin could be measured using a weighing balance or an analytical balance. Both are classified as weighing scales but the analytical balance can measure the mass of objects up to 4 decimal places, thus, providing better accuracy in measurement than ordinary weighing balance that can only measure up to 2 decimal places.
In order to measure 50 cm3 of the sea water, a graduated cylinder or a buret can be used. Both equipment can measure up to the same decimal places and, thus, have virtually the same accuracy.
Answer:
k = 4,92x10⁻³
Explanation:
For the reaction:
AB₂C (g) ⇄ B₂(g) + AC(g)
The equilibrium constant, k is defined as:
![k = \frac{[B_{2}][AC]}{[AB_2C]}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B%5BB_%7B2%7D%5D%5BAC%5D%7D%7B%5BAB_2C%5D%7D)
<em>(1)</em>
Molar concentration of the species are:
[AB₂C]: 0,0840mol / 5L = <em>0,0168M</em>
[B₂]: 0,0350mol / 5L = <em>0,0070M</em>
[AC]: 0,0590mol / 5L = <em>0,0118M</em>
Replacing this values in (1):
![k = \frac{[0,0070][0,0118]}{[0,0168]}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B%5B0%2C0070%5D%5B0%2C0118%5D%7D%7B%5B0%2C0168%5D%7D)
<em>k = 4,92x10⁻³</em>
I hope it helps!
