The equation in option A appears to be flawed.
The left hand side of the first equation contains eight carbon atoms whereas the right hand side contains only four. This equation is thus not properly balanced.
Answer:
The diagram above describes the movement of molecules in a substance used to make fuel. What change did the fuel scientist observe in the substance? (The diagram only shows some gray colored pots and underneath one it says before: molecules move around each other and the other one says after molecules move away from each other) I needd helpppp asappp A. Before the process, the substance was a solid. After the process, it was a liquid. B. Before the process, the substance was a gas. After the process, it was a liquid. C. Before the process, the substance was a liquid. After the process, it was a solid. D. Before the process, the substance was a liquid. After the process, it was a gas.
Let's see. It's definitely not the O horizon, because that layer is at the very top. A is already eliminated. Same thing with the A horizon, that is actually the top soil. C is out. It has to be either B or D. A D layer doesn't exist, so B is out. But the B horizon is considered a region of subsoil. The answer is D: B horizon.
Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives = 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = = = 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.
For example: Chlorine has 2 isotopes with the following abundances
Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%
Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %
Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =
= 26.4993 + 8.9531 = 35.4524 amu
Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average