Answer is: A) 124 s.
c₀ = 3 mol/L.
c₁ = 0,700 mol/L.
k = 8,8·10⁻³ 1/M·s.
Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t.
k·t = 1/0,700 - 1/3.
0,0088·t = 1,095.
t = 1,095 ÷ 0,0088.
t = 124 s.
c₀ - <span>initial concentration.
c</span>₁ - <span> concentration at a particular time.
k - </span><span>the rate constant.
t - time.</span>
Answer:
0.043 M
Explanation:
The reaction that takes place is:
- Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume required to reach the equivalence point</em>:
- 0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺
As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.
With that in mind we can <u>calculate the hydroxide ion concentration in the original sample solution</u>, using <em>the calculated number of moles and given volume</em>:
- 1.0817 mmol OH⁻ / 25.0 mL = 0.043 M
It is called a spring tide
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Answer:
The answer to your question is:
Explanation:
Data
carbon 7.3% = 7.3g
hydrogen 4.5% = 4.5g
oxygen 36.4% = 36.4 g
nitrogen 31.8% = 31.8 g
Now
For carbon
12 g --------------------1 mol
7.3 g ------------- x
x = 7.3/12 = 0.608 mol
For hydrogen
1 g -------------------- 1 mol
4.5 g ------------------ x
x = 4.5 mol
For oxygen
16 g ------------------- 1 mol
36.4 g ---------------- x
x = 2.28 mol
For nitrogen
14 g ---------------- 1 mol
31.8 g --------------- x
x = 2.27 mol
Now divide by the lowest result, the is 0.608 from carbon
carbon 0.608/0.608 = 1
hydrogen 4.5/ 0.608 = 7.4
oxygen 2.28/0.608 = 3.75
nitrogen 2.27/0.608 = 3.73
Empirical formula = CH₇O₄N₄
