Answer:
4.8 grams
Explanation:
Use PV=nRT
P: 775 mmHg (divide by 760 mmHg to get atm) -> 1.02 atm
V: 3700 mL (divide by 1000 to get L) -> 3.7L
n: ?
R: (a constant) 0.0821L * atm/k *mol
T: 33 C (add 273 to get K) -> 306K
Move equation so n is on the side: PV/RT = n. Plug the numbers into the equation.

Then, convert moles to grams using the molar mass of O2 which is 32g/mol.
= 4.8g
The net ionic equation of the reaction could be determined by cancelling out the like ions between both sides of the reaction. These ions are called spectator ions. They are called as such because they do not actively participate in the reaction. The spectator ions are Na+ and Cl-. When you cancel those, the equation would become letter D.
Answer:
D. number of neutrons
Explanation:
Isotopes of the same element must have the same number of neutrons. Neutrons of an atom are the uncharged particles within the nucleus.
Isotopy is the existence of two or more atoms of the same element having the same atomic number but having different mass number due to the differences in the number of neutrons in their various nuclei.
Therefore, isotopes differs based on the number of neutrons they contain.
Prevents oxidation of the metal
This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.
Then, we can write the following equilibrium expression for hydrofluoric acid once the change,
, has taken place:
![[HF]=[HF]_0-2x](https://tex.z-dn.net/?f=%5BHF%5D%3D%5BHF%5D_0-2x)
Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):
![8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }](https://tex.z-dn.net/?f=8.70x10%5E%7B-3%7D%3D%5Cfrac%7B%281.33x10%5E%7B-3%7D%20M%29%5E2%7D%7B%5BHF%5D%7D%20%7D)
![[HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M](https://tex.z-dn.net/?f=%5BHF%5D%3D%5Cfrac%7B%281.33x10%5E%7B-3%7D%20M%29%5E2%7D%7B8.70x10%5E%7B-3%7D%7D%20%7D%3D2.03x10%5E%7B-4%7DM)
Finally, the initial concentration of HF is calculated as follows:
![[HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M](https://tex.z-dn.net/?f=%5BHF%5D_0%3D%5BHF%5D%2B2x%3D2.033x10%5E%7B-4%7D%2B2%2A%281.33x10%5E%7B-3%7D%29%3D2.86x10%5E%7B-3%7DM)
Learn more: