Answer:
The correct name for P₄S₃ is tetraphosphorous trisulfide or also known as phosphorous sesquisulfide.
Explanation:
P represents the element phosphorous.
S represents the element sulfur.
Since there is a 4 beside P, then we are going to use the prefix "tetra-". Since there is a 3 beside S, then we are going to use the prefix "tri-". We will also use the suffix "-ide" in sulfur because we are naming a compound with multiple elements.
So, P₄S₃ will be named tetraphosphorous trisulfide. When using more scientific terms, then the name of the compound is phosphorous sesquisulfide.
Nature is full of wonders and one of the most impressive of them are symbiotic relationships (mutually beneficial relationships). Despite the competition in the kingdom of life, here 2 actors help each other to gain more than they could individually. Let's look at the answers. The sea anemone can't possibly fertilize the clownfish's eggs, only another clownfish can do that. The 4th choice might be correct but it does not help the clownfish; The 5th choice is also correct but it is not related to the offspring of the clownfish. For the 3rd choice, we have that the anemone sometimes provides leftovers for the fish, but the fish' eggs cannot use this source of nutrients. However, we know that the 1st assertion holds; the poisonous tentacles of the sea anemone protect the eggs from potential predators.
Answer:
Explanation:
A radioactive isotope is an isotope that undergoes nuclear decay, breaking apart into a smaller nucleus and emitting radiation during the process.
The half-life of an isotope is the amount of time it takes for a certain quantity of a radioactive isotope to halve.
For a radioactive isotope, the amount of substance left after a certain time t is:
(1)
where
is the mass of the substance at time t = 0
m(t) is the mass of the substance at time t
is the half-life of the isotope
In this problem, the isotope is uranium-235, which has a half-life of
We also know that the amount of uranium left in the rock sample is 6.25% of its original value, this means that
Substituting into (1) and solving for t, we can find how much time has passed:
Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g
Some patterns and trend that are present in the periodic table would be
1. electronegativity (from left-to-right it increases across the table)
2. ionization (from left-to right it increases and from bottom-to-top it increases)
3. electron affinity (same as ionization energy)
4. atom radius (increases opposite way; from right-to-left it increases and from top-to-bottom it increases)
5. melting point (higher melting points with metals and lower melting point with non-metals)
6. metallic character (same as atom radius)
