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2 years ago

Echoic memory and iconic memory are a part of the __________ memory.

Chemistry

2 answers:

Mekhanik [1.2K]2 years ago

7 0

Answer:

Sensory memory is a correct answer.

Explanation:

Echoic and iconic memories are a part of the sensory memory.

Sensory memory : where any form of information is sensed. Sensory memory store all the sensory information.

Sensory information is received by the sensory receptors and then it is sends to the nervous system and the information is processed.

Echoic memory is the type of sensory memory that is used for the auditory system. It is used to hold the information for about 3-4 seconds.

Iconic memory is the branch of the sensory memory, its memory for the visual stimuli is called iconic memory.

yulyashka [42]2 years ago

3 0

Echoic memory and iconic memory are a part of the Sensory memory.

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A ball-and-stick model of a molecule of formaldehyde, ch2o, is made from the following components: what do the sticks represent?

9966 [12]

In chemistry, the ball-and-stick model is a molecular ideal of a chemical matter which is to expose both the three-dimensional position of the atoms and the bonds among them. The atoms are normally symbolise by spheres, join by rods which shows the bonds.

Formaldehyde forms formaldehyde structure bond it shares double bond with O2 atoms.

Formaldehyde also known as methanol .

Methanol is colourless.

It is flammable.

It is gas at room temperature.

Methanol having pungent odor and it is a volatile organic compounds.

It is made by the composition of Hydrogen, oxygen, and carbons.

To know more about Formaldehyde here :

brainly.com/question/14895085?referrer=searchResults

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1 year ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

3 years ago

What is the relationship between atoms and static electricity

Mnenie [13.5K]

They both build up to form electricity

3 0

2 years ago

Read 2 more answers

Antoine Lavoisier burned metals in sealed jars. What is your prediction of his experimental results?

Aliun [14]

Answer:

The mass of the jar and contents remained the same after the metal was burned.

Explanation:

My prediction about the experimental results is that the mass of the jar and contents remained the same after the metal was burned in the jar.

This is compliance with the law of conservation of mass which states that in a chemical reaction, matter is neither created nor destroyed by bonds are rearranged for new compounds to form.

In compliance with this law, it is expected that the mass of the jar and its content will remain the same before and after the reaction.
No new material was added and no material was removed from the jar.

8 0

2 years ago

Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is

olga nikolaevna [1]

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid $H_3O^{+}$ is shown in the image.

m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.

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3 years ago