not sure but h = hydrogen and o = oxygen
Answer:
The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.
Explanation:
Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.
Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).
As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.
As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.
The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.
Force = mass x gravity
Force = 20 kg x 9.8 m/s²
Force = 196 Newtons
Answer A
hope this helps!
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
