Answer:
c. 0.2 M HNO₃ and 0.4 M NaF
.
Explanation:
A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.
A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.
Thus:
a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. <em>IS NOT </em>a buffer.
b. 0.2 M HNO₃ and 0.4 M HF
. This is a mixture of two strong acids. <em>IS NOT </em>a buffer.
c. 0.2 M HNO₃ and 0.4 M NaF
. NaF is the conjugate base of a weak acid as HF is.
The reaction of HNO₃ with NaF is:
HNO₃ + NaF → HF + NaNO₃
That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture <em>IS </em>a buffer.
d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this <em>IS NOT </em>a buffer.
I hope it helps!
Answer:
1/4 or 25%
Explanation:
The Arctic region of the earth refers to that part of the earth around the north pole region. Hence, when we are talking about latitude O degrees North, the areas around this geographical location is referred to as the arctic.
Now , there is an estimated 1/4 or 25% of the world’s oil and natural gas here. Unfortunately, these are yet accessible because of the amount of ice or snow covering. With increase in technological advancements, this might be accessible in the nearest future
Uranium can be mined and then immediately used in a nuclear power plant.
False
It has to have chemicals put onto it before it can be put on or else it will explode
Answer:
150mL
Explanation:
First, let us calculate the volume of the diluted solution. This is illustrated below:
Information obtained from the question include:
V1 (initial volume) = 50mL
C1 (initial concentration) = 1M
C2 (final concentration) = 0.25M
V2 (final volume) =.?
The volume of the diluted solution can be obtained using the dilution formula C1V1 = C2V2 as shown below:
C1V1 = C2V2
1 x 50 = 0.25 x V2
Divide both side by 0.25
V2 = 50/0.25
V2 = 200mL
The volume of the diluted solution is 200mL.
Now, to obtain the volume of the water added to the solution, we'll simply subtract the initial volume from the final volume as shown below
Volume of water added = V2 - V1
Volume of water added = 200 - 50
Volume of water added = 150mL
Therefore, the volume of water added is 150mL
