Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on matter.
Hi
I think it’s B
Hopefully this helps!!
<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 8.04 days
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 39 days
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the sample = 8.0 grams
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}](https://tex.z-dn.net/?f=0.0862%3D%5Cfrac%7B2.303%7D%7B39%7D%5Clog%5Cfrac%7B8.0%7D%7B%5BA%5D%7D)
![[A]=0.278g](https://tex.z-dn.net/?f=%5BA%5D%3D0.278g)
Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams
True.
without natural selection darwin’s theory would not make sense :))
Based on the calculations, the mass of Aluminum in 4.85 × 10²² atoms is equal to 2.1762 grams.
<h3>How to calculate the mass of Aluminum?</h3>
In order to determine the mass of Aluminum, we would calculate the number of atoms in 1 mole of an Aluminum atom in accordance with Avogadro's constant.
1 mole of Aluminum atom = 6.02 × 10²³ molecules
X moles of Aluminum atom = 4.85 × 10²² molecules
Cross-multiplying, we have:
X = 4.85 × 10²²/6.02 × 10²³
X = 0.0806 moles.
Mass = Molar mass × Number of moles
Mass = 27 × 0.0806
Mass = 2.1762 grams.
Read more on moles here: brainly.com/question/3173452
#SPJ1
