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3 years ago

Match each type of learning to the correct definition

Chemistry

2 answers:

Radda [10]3 years ago

7 0

Answer:

Insight learning=2 one

conditioning=4th one

Imprinting=1st one

Trial-and-error learning=3rd one

(If urs is not in the same order here is the written one)

Insight learning-the process of using prior knowledge to solve a problem

Conditioning-the process is associating a stimulus with a good or bad outcome

Imprinting-a bond an individual or object shortly after birth or hatching

Trial-and-error learning-the process of performing a behavior more and more skillfully.

Explanation: hope it helps got it right on edu 2021

pishuonlain [190]3 years ago

3 0

Answer:

conditioning-the process of associating a stimulus with a good or bad outcome.

trial-and-error learning-the process of performing a behavior more and more skillfully.

insight learning-the process of using prior knowledge to solve a problem.

imprinting-a bond to an indivisual or object shortly after birth or hatching.

Explanation:

Did this assignment.

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Which of these is the final result of secondadry succession?

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If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

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3 years ago

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3 years ago

How does weathering change rocks and minerals

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3 years ago

a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what

e-lub [12.9K]

The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
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p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa

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3 years ago