The answer:
for the monoatomic <span>selenium ions
</span> -the ion charge of selenium is 2-, so the answer is [Se]2+
as for the monoatomic phosphorus ions
-the ion charge of phosphorus is 3-, so the answer is [P]3-
Answer:
The collision theory states that a chemical reaction can only occur between particles when they collide (hit each other).
<em>h</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em><em>~</em>
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
Learn more: brainly.com/question/16243729
2NaClO3 = 2NaCl + 3O2
Hopes this helps <33
Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old
