Did you intend to write [PdCl4]^-2 instead of PdCl2-4? If so, then:
<span>Cathode: [PdCl4]^-2(aq) + 2e- ======⇒ Pd(s) + 4Cl-(aq) </span>
<span>Anode: Cd(s) ==⇒ Cd+2(aq) + 2e-</span>
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
4. Blue, Red, Green, Purple
(Lowest densities on the top, highest densities on the bottom)
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
Answer: The concentration of the reactants and products remain constant!
Explanation: <em>remember that </em><em>a dynamic equilibrium is a state of balance in which the rates of the forward and reverse reactions are equal,</em><em> and they continue to occur at an equivalent and constant rate with the same amount of product and reactants!</em>
