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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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2 years ago

When Wolverine’s 10-pound adamantium claws are dissolved in 100 mL of 10 M nitric acid, 10.7 grams of adamantium nitrate are rec

Mazyrski [523]

The percent yield is 71.3 %.

Explanation:

Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.

$Percent Yield = \frac{Experimental Outcome}{Theoretical Outcome} *100$

So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.

Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.

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2 years ago

PLS HELP THE QUESTION IS ON THE PICTURE

IceJOKER [234]

Concepts used:

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: Number of formula units = number of moles * 6.022*10²³

number of moles of an element or a compound = given mass/molar mass

__________________________________________________________

003 - Number of CaH₂ formula units in 6.065 grams

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

004 - Mass of 6.34 * 10²⁴ formula units of NaBF₄

Number of Moles:

We mentioned this formula before:

Number of formula units = number of moles * 6.022*10²³

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles= 10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

005 - Number of moles in 9.78 * 10²¹ formula units of CeI₃

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles OR 0.016 moles

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2 years ago