Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
Explanation:
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
Answer:
[H⁺] = 1.58x10⁻⁶M; [OH⁻] = 6.31x10⁻⁹M.
pH = 8.23; pOH = 5.77
Explanation:
pH is defined as <em>-log [H⁺]</em> and also you have <em>14 = pH + pOH </em>
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Thus, for a solution of pH = 5.80.
5.80 = -log [H⁺] → [H⁺] = 10^-(5.80) = 1.58x10⁻⁶M
pOH = 14-5.80 = 8.20 → [OH⁻] = 10^-(8.20) = 6.31x10⁻⁹M
Thus, for a solution of [H⁺] = 5.90x10⁻⁹M and pH = -log 5.90x10⁻⁹M = 8.23
And pOH = 14-8.23 = 5.77
Explanation:
Gay-Lussac's law states that if P=pressure and T=temperature, then P/T=P/T. Temperature must be in Kelvin, but pressure can be in any unit.
To convert Celsius into Kelvin, just add 273: 10+273=283K, and 28+273=301K.
So using the equation,
200kPa÷283=? kPa÷301
Solving this equation gives you about 212.72 kPa.
G(2)=2
For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.
