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olga nikolaevna [1]
3 years ago
10

Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest

amount. The threshold limit is 2.0 x 10^-11 g per liter of air. If the current price of 50 g of vanillin is $112, determine the cost to supply enough vanillin so that the aroma could be detected in a large aircraft hangar with a volume of 5.0 x 10^7 ft^3.
Chemistry
1 answer:
balu736 [363]3 years ago
5 0

Answer:

Cost to supply enough vanillin is 3.2\$

Explanation:

Threshold limit of vanillin in air is 2.0\times 10^{-11}g per litre means there should be 2.0\times 10^{-11}g of vanillin in 1L of air to detect aroma of vanillin.

1ft^{3}=28.32L

So, 5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L

So amount of vanillin should be present to detect = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g

As cost of 50 g vanillin is 112\$ therefore cost of  (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)gvanillin = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$

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GalinKa [24]

Answer is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.

<span> Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] = 3s(Ca₃(PO₄)₂) = 3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.

s = ⁵√(Ksp ÷ 108).

6 0
3 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-cell reactions are:

Oxidation half reaction (anode):  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Pb^{2+}+2e^-\rightarrow Pb

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

Therefore, the cell potential for this reaction is 0.50 V

5 0
4 years ago
Solid potassium chlorate decomposes upon heating to form
olga55 [171]

Answer:

32.6%

Explanation:

Equation of reaction

2KClO₃ (s) → 2KCl (s) + 3O₂ (g)

Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)

Molar volume of Oxygen at s.t.p = 22.4L / mol

since the gas was collected over water,

total pressure = pressure of water vapor + pressure of  oxygen gas

0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C

pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1

P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p

Using ideal gas equation

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = \frac{245.2 * 0.66411}{67.2}  = 2.4232 g

percentage of potassium chlorate in the original mixture = \frac{2.4232 * 100}{7.44} = 32.6%

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3 years ago
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The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

7 0
3 years ago
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