Answer:
Oxygen with 0.36 moles left over
Explanation:
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
b
Explanation:
February 7, 1863, was the day John Newlands published a paper outlining what would be known as “The Law of Octaves”. Newlands discovered if he ordered the known elements by increasing atomic weight, the chemical properties of the elements would be similar for every eighth group
Answer:
A lower ph is always more acidic, due to the increased concentration of hydrogen ions in the solution/substance.
A ph of 3 is 100 times more acidic than a pH of 5, and this is due to the increments on the scale.
Answer:
Specific heat of metal = 0.26 j/g.°C
Explanation:
Given data:
Mass of sample = 80.0 g
Initial temperature = 55.5 °C
Final temperature = 81.75 °C
Amount of heat absorbed = 540 j
Specific heat of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 81.75 °C - 55.5 °C
ΔT = 26.25 °C
540 j = 80 g × c × 26.25 °C
540 j = 2100 g.°C× c
540 j / 2100 g.°C = c
c = 0.26 j/g.°C
