Answer:
A) t = 22.5 min and B) t = 29.94 min
Explanation:
Initial concentration, [A]₀ = 100
Final concentration = 100 -75 = 25
Time = 45 min
A) First order reaction
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[25] − ln[100] = - 45k
-1.386 = -45k
k = 0.0308 min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
ln[A] − ln[A]₀ = −kt
Solving for k;
ln[50] − ln[100] = - 0.0308 * t
-0.693 = -0.0308 * t
t = 22.5 min
B) Zero Order
[A] = [A]₀ − kt
Using the values from the initial reaction and solving for k, we have;
25 = 100 - k(45)
-75 = -45k
k = 1.67 M min-1
How long after its start will the reaction be 50% complete?
Initial concentration, [A]₀ = 100
Final concentration, [A] = 100 -50 = 50
Time = ?
[A] = [A]₀ − kt
50 = 100 - (1.67)t
-50 = - 1.67t
t = 29.94 min
Vanadium (V)
Vanadium is the only one in the 4th period here so
The density of the block will be "7.11 gm/cm³". A further explanation is provided below.
Given values are:
Density of iron,
Mass of block of metal,
Volume of block,
By using the formula,
→ 
then,
→ The density of block will be:
= 
= 
= 
Learn more:
brainly.com/question/10804879
Answer:
Newton's third law of motion states that every action, there is an equal and opposite reaction force and that forces come in pairs
Answer:
Answer is given below.
Explanation:
Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.
In cell representation, half cell present left to salt-bridge notation
is anodic system and another half cell present right to salt-bridge notation
is cathodic system.
So anode is Cu and cathode is Ag.
oxidation: 
[reduction:
]
-----------------------------------------------------------------------------------------------
chemical equation: 
Oxidizing agent is that species which takes electron from another species. Here
takes electron from Cu. Hence
is the oxidizing agent.
Reducing agent is that species which gives electron to another species. Here Cu gives electron to
. Hence Cu is the reducing agent.