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makvit [3.9K]
2 years ago
4

What process did you use to get Ricky the Lab Rodent to repeat the activity assigned? (Site 1)

Chemistry
1 answer:
pogonyaev2 years ago
8 0

Answer:

Operant conditioning was used to influence the behavior of the rodent. By using reward tactics for the response we wanted, it influenced the actions of the rodent, making him complete the wanted action more than the others. The rodent was conditioned to complete a certain task, then was rewarded with cheese.

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why is it less effective to wash an insoluble precipitate with 15 ml of water once than it is to wash the precipitate with 3 ml
MArishka [77]

It is less effective to wash an insoluble precipitate with 15 ml of water once than it is to wash the precipitate with 3 ml of water 5 times because commonly, when you clean an <span>indissoluble precipitate with water, the water will not be completely saturated with contaminates. Therefore, the absorption of the contaminates would lower with each wash, since if you only washed it once with a bigger amount or volume of water, it’d become less contaminated with the wash water but it wouldn’t get rinsed numerous times.</span>

6 0
2 years ago
For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?You do not need to look up any values to answer t
Nutka1998 [239]

Answer:

Reactions 1, 3 and 5

Explanation:

First thing's first, let's ensure that all the reactions given are balanced. This is given as;

CO(g) + 1/2 O2(g )→ CO2(g)

Li(s) + 1/2 F2(l) → LiF(s)

C(s) + O2(g) → CO2(g)

CaCO3(g) → CaO + CO2(g)

2Li(s) + F2(g) → 2LiF(s)

For the condition to be valid;

- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.

- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.

This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.

3 0
2 years ago
Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products
hram777 [196]

<u>Answer:</u> This illustrates law of conservation of mass.

<u>Explanation:</u>

Dalton's theory is based on mainly two laws which are law of conservation of mass and law of constant proportion.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of calcium carbonate follows:

CaCO_3\rightarrow CaO+CO_2

We are given:

Mass of calcium carbonate = 100 grams

Mass of calcium oxide = 56 grams

Mass of carbon dioxide = 44 grams

Total mass on reactant side = 100  g

Total mass on product side = 56 + 44 = 100 g

As, the total mass on reactant side is equal to the total mass on product side.

Thus, this illustrates law of conservation of mass.

6 0
2 years ago
The graph shows the amount of a gaseous product formed over time during two trials of a reaction. A different concentration of a
ss7ja [257]

Answer:

Trial 2, because the amount of product formed per unit time is higher.

Explanation:

Message me for extra help.

8 0
2 years ago
Read 2 more answers
Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0
2 years ago
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