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3 years ago

When a response to information is initiated, the nervous system carries the response to the rest of the body.

2 answers:

6 0

Technically, this statement is false. When a response to information is initiated, the PERIPHERAL NERVOUS SYSTEM carries the response to the rest of the body. It would be different saying NERVOUS SYSTEM since there are two; CENTRAL AND PERIPHERAL. Since Central initiates and integrates information and appropriate responses while the peripheral carries those responses.

zvonat [6]3 years ago

3 0

First:

This is not a true or false question.

Second:

The answer is Peripheral.

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1.Oxygen (O) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and the other elements

Lady_Fox [76]

1. A. All the elements in the column have similar chemical properties.

2. Substances on the periodic table cannot be broken down into other substances and are therefore elements.

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3 years ago

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If the rock has a mass of 250.8 grams, what is its density?

cluponka [151]

To know the density you also need to know the volume of the rock.

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3 years ago

A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti

MrRa [10]

Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

$H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_3H_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL$

Putting values in above equation, we get:

$2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M$

Thus the molarity of the malonic acid solution is 0.08335 M

5 0

3 years ago

Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its stan

LenKa [72]

solution:

$the reaction for formation of methyl acetate is\\CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\standard internal energy \\\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\where \delta n=change in number of moles=0\\\delta u=\delta H\\=-442kj/mol$

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3 years ago

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What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?

jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

Moles of Aluminium is 1.40 mol
Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

= 1.4 moles × 0.5

= 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

= 71.372 g

3 0

3 years ago