Answer:
Thus, the order of the reaction is 2.
The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹
Explanation:
The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.
The concentration vs time graph of zero order reactions is linear with negative slope.
The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.
The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.
Considering the question,
A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.
<u>Thus, the order of the reaction is 2.</u>
<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>
The correct answer is C. It was based on the thoughts of an early philosopher.
Explanation:
An atom is a basic and smallest unit that composes matter and that determine the properties of elements. Regarding the development of ideas related to atoms these did not begin in science but in philosophy; indeed the first person that proposed matter or elements were composed of certain smaller units was the philosophers Leucippus and his pupil Democritus in Ancient Greece, who stated atoms were eternal, infinite and defined the qualities of an object, idea that was supported by other Greek philosophers. But it was not until 16th and 17th centuries after the Middle Ages that the term re-emerged and until the 19th century it was officially proposed and there were experiments by scientists that later became a theory. Therefore, the development of the earliest idea about attoms differs from later work of scientists is that the earliest idea was based on the thoughts of an early philosopher.
No, there is a rule called HONC... they could also bond with o2, n, and c
<h2>Solutions:</h2>
<u>Case a:</u> Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.75 x 10⁻⁵]
pH = 4.75
<u>Case b:</u> Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [6.50 × 10⁻¹⁰]
pH = 9.18
<u>Case c:</u> Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.0 × 10⁻⁴]
pH = 4
<u>Case d:</u> Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.50 × 10⁻⁵]
pH = 4.82
