Answer: Volume of KCl is
Explanation: For the given reaction,
We can see that 2 moles of is giving 2 moles of KCl. So, the number of moles of KCl will be 2.
Mass of KCl can be calculated by,
Molar mass of KCl = 74.5 g/mol.
Number of Moles = 2
Given mass = 149g
Now, to calculate the Volume of KCl, we will use the density formula,
Mass = 149g (Calculated above)
Putting the values in density formula, we get
Conversion Factor:
Volume of KCl in
The given question is incomplete. The complete question is :
The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: The half-life of the reaction is 35.0 s at 680°C. Calculate the first order rate constant.
Answer: a) The first order rate constant is
b) The time after which 95% reactions gets completed is 151 seconds
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for finding the rate constant
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
The first order rate constant is
b) for completion of 95 % of reaction
The time after which 95% reactions gets completed is 151 seconds
Answer:
Here's what I get
Explanation:
You haven't shown your drawing, so I will assume that Metal A is the anode and Metal B is the cathode.
I will make a galvanic cell using Mg and Zn as the metals.
I selected Zn because it is common and readily available in the lab.
Zn is lower than Mg in the activity series, so Mg should be able to displace Zn from its salts
The standard reduction potentials are:
<u> E°/V </u>
Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s); -0.76
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s); -2.38
The Mg half-reaction has the more negative potential, so it will be the oxidation half-reaction.
8 and 9. Oxidation and reduction half-reactions and cell potential
<u> E°/V</u>
Oxidation: Mg(s) ⇌ Mg²⁺(aq) + 2e⁻ ; +2.38
<u>Reduction</u><u>: Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s);</u> <u>-0.76</u>
Mg(s) + Zn²⁺ (aq) ⇌ Mg²⁺(aq) + Zn(s); +1.62
The cell potential is positive, so the reaction will be spontaneous.
Mg is the anode, so it is Metal A.
Zn is the cathode, so it is Metal B.
1. The Mg|Mg²⁺ half-cell is the oxidation compartment.
2. The Zn²⁺|Zn half-cell is the reduction compartment.
3. The electrons flow from anode to cathode in the external circuit.
4. The Mg²⁺ ions flow from the Mg through the solution to the salt bridge.
5. The Zn²⁺ ions flow from the solution to the Zn.
6. NO₃⁻ ions flow from the salt bridge into the anode compartment to balance the charge of the developing Zn²⁺ ions.
7. Na⁺ ions flow from the salt bridge into the cathode compartment to replace the charge of the depleted Zn²⁺ ions.
