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4 years ago

Conditioning is the process of

1 answer:

7 0

Conditioning is the process of C. learning associations.

Two types of conditioning.
1) Classical conditioning
2) Operant conditioning.

Classical conditioning happens when an individual learns to associate two stimuli. This happens when an association is learned between an unconditioned stimulus and a conditioned stimulus.

Operant conditioning happens when an individual associates a behaviour with a consequence. For example, when a behaviour produces a good consequence; the individual is conditioned to repeat the behaviour to garner the same good consequence, and vice versa.

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Give an example of oxidation reaction which can be a nuisance

babunello [35]

Magnesium oxide can be very bad for your health, and when we did an experiment with it in class it was white because it was so hot. It is very flammable.

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3 years ago

One success of Johann Döbereiner's law of triads was that it organized most of the known elements into three-element groups base

jenyasd209 [6]

That statement is true

it was easy to organize 'all' the elements in a group of threes because back then they only knew about 15 - 20 elements

hope this helps

4 0

3 years ago

Read 2 more answers

At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L

alexira [117]

We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K

5 0

3 years ago

Read 2 more answers

How many liters of a 0.30M solution are needed to give 2.7 moles of solute? Your answer:

OverLord2011 [107]

You HAVE TO know that molarity (M) tells you the number of moles of a solute per Liters of solution.
M=mol/L
0.3M = 2.7 mol/ Volume
Volume = 2.7 mol/3.0 L
Volume = 0.9 L

8 0

4 years ago

Read 2 more answers

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

4 years ago