Answer:
The correct answer is: 1.035 x 10⁻³ M
Explanation:
The dissociation equilibrium for acetic acid (CH₃COOH) is the following:
CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵
The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):
![Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7D%20COO%5E%7B-%7D%20%5D%5BH%5E%7B%2B%7D%20%5D%7D%7B%5BCH_%7B3%7D%20COOH%5D%7D%3D%201.8%20x%2010%5E%7B-5%7D)
Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:
![\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.016%20M%29%5BH%5E%7B%2B%7D%20%5D%7D%7B%280.92M%29%7D%3D%201.8%20x%2010%5E%7B-5%7D)
⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M
Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Polymerization
For example:
nCH₂=CH₂ ---p,T,kat.----> -[-CH₂-CH₂-]- n
Answer:
the answer would be a i did the test
Explanation: