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laiz [17]
3 years ago
6

What nevres carrys information from the hairs inside the cochlea?

Chemistry
1 answer:
Slav-nsk [51]3 years ago
5 0
The cochlear nerve(also auditory or acoustic nerve)
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Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh
Ann [662]

Answer:

See explanation and image attached

Explanation:

The  Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi.  R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the  lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

7 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
Some one please help me ​
Norma-Jean [14]

Answer:

C.4

Explanation:

6 0
3 years ago
An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well Exponents in rate laws are
Karo-lina-s [1.5K]

Answer:

- False.

- False.

- True.

- True.

Explanation:

Hello, for each statement we state:

- An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well.

FALSE because considering a rate law like:

-r=kC^2

The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

-r=kC

The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

- Exponents in rate laws are based on the coefficients from the balanced equation.

FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

- The rate constant, k, takes into account the effect of activation energy and temperature on the reaction.

TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

k=Aexp(-\frac{Ea}{RT})

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...

Best regards.

5 0
3 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5
NNADVOKAT [17]

For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as

T2=181.55°C\

<h3> What temperature (in °C) would the vapor pressure be 0.500 atm? </h3>

Generally, the equation for the gas  is mathematically given as

ln(p1/p2)=dHvap/R(1/T2-1/T1)

Therefore

ln(1/0.26)=23500/8.214(1/T2-1/555)

T2=181.55^C

In conclusion

T2=181.55°C

Read more about Temperature

brainly.com/question/13439286

7 0
2 years ago
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