Answer:
runoff is he draining away of water (or substances carried in it) from the surface of an area of land, a building or structure, etc.
Explanation:
Answer:
A. there is an isotope of lanthanum with an atomic mass of 138.9
Explanation:
By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.
Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.
Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.
Thus, the correct option is:
<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>
The question is missing the graphics required to answer which I have attached as an image.
There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.
We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.
The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.
The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
The scientist observes at what rate is the concentration increasing or decreasing.