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3 years ago

From what you've learned so far, how is molecular structure related to smell?

Chemistry

1 answer:

Sidana [21]3 years ago

7 0

Answer:

Changing the shape of the molecules that create fragrances in a flower or fruit may influence our perception of their smell. The reaction pattern produced, olfactory code, is sent as a signal to the brain, which which is how you smell things.

Explanation:

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Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5

Anuta_ua [19.1K]

Explanation:

pH is the negative logarithm of hydronium ion concentration present in a solution.

If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

$pH=-\log[H_3O^+]$ ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14 ........(2)

For 1:

We are given:

pH = 5.54

Putting values in equation 1, we get:

$5.54=-\log[H_3O^+]$

$[H_3O^+]=2.88\times 10^{-6}M$

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

For 2:

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

$4.3=-\log[H_3O^+]$

$[H_3O^+]=5.012\times 10^{-5}M$

The solution is acidic in nature.

For 3:

We are given:

pH = 7.0

Putting values in equation 1, we get:

$7.0=-\log[H_3O^+]$

$[H_3O^+]=1.00\times 10^{-7}M$

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

For 4:

We are given:

pH = 12.9

Putting values in equation 1, we get:

$12.9=-\log[H_3O^+]$

$[H_3O^+]=1.26\times 10^{-13}M$

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

For 5:

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

$12.8=-\log[H_3O^+]$

$[H_3O^+]=1.58\times 10^{-13}M$

The solution is basic in nature.

For 6:

We are given:

$[H_3O^+]=1\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(1\times 10^{-5})$

$pH=5$

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0

3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

3 years ago

An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.

____ [38]

Answer:

(a) Density of the air = 1.204 kg/m3

(b) Pressure = 93772 Pa or 0.703 mmHg

Explanation:

(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

$\rho_{air}=\frac{P}{R*T}$

First, we change the units of P to Pa and T to deg K:

$P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K$

Then we have

$\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}$

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

$PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg$

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.

$F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N$