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3 years ago

Why do different substances have different Odors?

Chemistry

1 answer:

mixer [17]3 years ago

4 0

Answer:

Substances generate a smell when their molecules land on so-called olfactory neurones in our noses (which, for some things, is a pretty unpleasant thought). ... But this fails to explain why some molecules with similar shapes can smell completely different, while others with quite different shapes can have a similar scent.

Explanation:

I took chemistry

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Which elements in group is the most reactive nonmetals

klemol [59]

Answer:

flourine

Explanation:

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1 year ago

18.A 2.50 L sample of dry air in a cylinder exerts a pressure of

Vlada [557]

Answer:

$\boxed{\text{7.50 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{3.00 atm}\qquad & V_{1} &= & \text{2.50 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}3.00 \times 2.50 & =& 1.00V_{2}\\7.500 & = & 1.00V_{2}\\V_{2} & = &\textbf{7.50 L}\\\end{array}\\\text{The volume of the gas is } \boxed{\textbf{7.50 L}}$

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3 years ago

Find the place of the digit 1 in the number 81389398.

dlinn [17]

Answer:

millions

Explanation:

the 1 is seven decimbal places to the left (seven places left of zero), which is the millions place

5 0

3 years ago

Read 2 more answers

Consider an acidic chromate-dichromate system at equilibrium in which the color is an orange-yellow. The reaction is represented

goldenfox [79]

Answer:

The equilibrium will move in forward direction.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

$Cr_2O_7^{2-}(aq)+H_2O(l)\rightleftharpoons 2CrO_4^{+}(aq)+2H^+(aq)$

On addition of base at the equilibrium, the hydroxide ions of the base will neutralize the hydrogen ions and lowering in concentration of hydrogen ion will be observed.

So, on lowering of concentration of hydrogen ions the equilibrium will move in direction in accordance to Le Chatelier’s Principle .The equilibrium will move in forward direction.

4 0

3 years ago

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volu

11111nata11111 [884]

Answer:

(a) 13.64; (b) 8.04; (c) 2.25

Explanation:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

$K_{\text{sp}} = {\text{[Ag$^{+}$][I$^{-}$]} = 8.3\times 10^{-17}$

(a) pAg at 35.10 mL

$\text{Moles of I$^{-}$} = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag$^{+}$} = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 1.822 × 10⁻³ 2.040 × 10⁻³

C/mol: -1.822 × 10⁻³ -1.822 × 10⁻³

E/mol: 0 0.218 × 10⁻³

We have a saturated solution of AgI containing 0.218 × 10⁻³ mol of excess I⁻.

V = 25.00 mL + 35.10 mL = 60.10 mL

$\text{[I$^{-}$]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}\\$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s 3.57 × 10⁻³ + s

$K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}\\$

Check for negligibility:

$\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag$^{+}$]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}$

(b) At equilibrium

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s s

$K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}$

(c) At 47.10 mL

$\text{Moles of Ag$^{+}$} = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 2.444 × 10⁻³ 2.040 × 10⁻³

C/mol: -2.040 × 10⁻³ -2.040 × 10⁻³

E/mol: 0.404 × 10⁻³ 0

V = 25.00 mL + 47.10 mL = 72.10 mL

$\text{[Ag$^{+}$]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}$

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4 years ago