Answer:
1) Carbon-14 is a naturally occurring carbon isotope.
2) Carbon-14 has a known half-life.
3) Carbon-14 is present in all living organisms.
Living things have carbon-14 in their organism, but when they died, the amount of C-14 they contain begins to decrease as the C-14 undergoes radioactive decay. Measuring the amount of C-14 in a sample from a dead plant or animal when the animal or plant died.
The half-life of C-14 is about 5730 years.
Half-life is the time required for a quantity (in this example number of radioactive nuclei of carbon-14) to reduce to half its initial value.
Hope this helps and also this question was already on Brainly check it out I just copied lol :)
The grams of aluminium oxide are formed when 350 kj heat are released is calculated as follows
1mole = 850Kj
what about 350kj
=1mole x350/850 = 0.412 moles
mass of Al = moles of Al x molar mass of Al
= 0.412mol x 27 g/mol = 11.124 grams
Answer:
The gain of an electron adds more electrons to the outermost shell which increases the radius because there are now more electrons further away from the nucleus and there are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an increase in atomic radius.GOOD LESSONS ♡
Explanation:
The given data is as follows.
= 9, 
= ?
Z for hydrogen = 1
As we know that,
Energy (E) = 
where, h = planck's constant = 
Js
c = speed of light = 
m/s
= wavelength
According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.
![\Delta E = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%28Z%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B1%7D%7D%5D)
or, = ![-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]](https://tex.z-dn.net/?f=-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%28Z%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B1%7D%7D%5D)
Putting the given values into the above equation as follows.
=
= ![-2.179 \times 10^{-18} J \times (1)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(9)^{2}}]](https://tex.z-dn.net/?f=-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%281%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7B%289%29%5E%7B2%7D%7D%5D)
n = 2
Thus, we can conclude that the final level of the electron is 2.
Explanation:
Molar mass of CaCl2 = 110.98g/mol
Moles of CaCl2 = 30.0 / 110.98 = 0.270mol
Moles of AgCl = 0.270mol * 2 = 0.540mol
Mass of AgCl = 0.540mol * (143.32g/mol) = 77.39g.
