Answer:
energy goes into the chemical and cool downs the surrounding area.
Answer:
1. Reflex Conditioning
Explanation:
Conditioning is an aspect of learning where a stimulus works effectively in producing a response from an organism. This response becomes regular given the type of reinforcement that is administered to the organism. The reinforcement is usually a reward that is given to the organism.
This is what is observed in the woodpecker. The stimulus or reinforcement which proves effective in producing a continuous response from the woodpecker is the abundant supply of the birds favorite bugs. This reinforcement makes it possible for the woodpecker to become conditioned towards returning to the tree.
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M
Answer:
3.10×10¯⁵ ft³.
Explanation:
The following data were obtained from the question:
Density (D) of lead = 11.4 g/cm³
Mass (m) of lead = 10 g
Volume (V) of lead =.?
Density (D) = mass (m) / Volume (V)
D = m/V
11.4 = 10 / V
Cross multiply
11.4 × V = 10
Divide both side by 11.4
V = 10 / 11.4
V = 0.877 cm³
Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:
1 cm³ = 3.531×10¯⁵ ft³
Therefore,
0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³
0.877 cm³ = 3.10×10¯⁵ ft³
Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.
Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.
