Answer:
2 H⁺ + 2e = H₂ ( reduction )
Explanation:
Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂
Fe( s ) = Fe⁺² + 2e ( oxidation )
2 H⁺ + 2e = H₂ ( reduction )
Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.
Ca + 2HCl = CaCl₂ + H₂
c=4.50 mol/l
v=2.20 l
n(HCl)=cv
m(Ca)/M(Ca)=n(HCl)/2
m(Ca)=M(Ca)cv/2
m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g
198 grams of Ca are needed
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Hope this helped.
Protons and neutrons
hope this helps
