Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O
Answer:
Balancing the equation
2KMnO₂+10KCl+8H₂SO₄⇒2MnSO₄+6K₂SO₄+8H₂O+5Cl₂
Answer:
A = 1,13x10¹⁰
Ea = 16,7 kJ/mol
Explanation:
Using Arrhenius law:
ln k = -Ea/R × 1/T + ln(A)
You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).
Using the values you will obtain:
y = -2006,9 x +23,147
As R = 8,314472x10⁻³ kJ/molK:
-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹
<em>Ea = 16,7 kJ/mol</em>
Pre-exponential factor is:
ln A = 23,147
A = e^23,147
<em>A = 1,13x10¹⁰</em>
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I hope it helps!
Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
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