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Katarina [22]
2 years ago
10

Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it

undergoes a first-order breakdown with a half-life of 9.0 min. An industrial flavor-enhancing process requires that a butter-flavored food be heated briefly at 200°C. How long can the food be heated at this temperature and retain 74% of its buttery flavor?
Chemistry
1 answer:
Sholpan [36]2 years ago
4 0

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

t__{1/2}}= \frac{0.693}{k}

where;

k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

k = \frac{0.693}{9.0 min}

k = 0.077 min^{-1}

Let the initial amount of butter flavor in the food be (N_0) = 100%

Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

t = - (\frac{1}{0.077}*In\frac{74}{100}  )

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

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a square boat made from iron has overall dimensions of 2.00 cm x 11.0 cm x 11.0 cm. it has a mass of 213 g. what is its density
Rus_ich [418]
Volume:

2.00 x 11.0 x 11.0 => 242 cm³

mass : 213 g

D = m / V

D = 213 / 242

D = 0.880 g/cm³

Answer B

hope this helps!
5 0
2 years ago
Read 2 more answers
What is the molarity of a solution made by dissolving 68.4g of NaOH in enough water to produce a 875 ml solution?
PilotLPTM [1.2K]
We first calculate for the number of moles of NaOH by dividing the given mass by the molar mass of NaOH which is equal to 40 g/mol. Solving,
                   moles of NaOH = (68.4 g/ 40 g/mol) = 1.71 moles NaOH
Then, we divide the calculate number of moles by the volume in liters. 
                     molarity = (1.71 moles NaOH / 0.875 L solution)
                       molarity = 1.95 M
8 0
1 year ago
Why does 4.03/0.0000035 = 1.2 x 106, instead of a different number of significant figures?
jasenka [17]

Explanations:- As per the significant figures rule, In multiplication and division, we go with least number of sig figs.

4.03 has three sig figs where as 0.0000035 has two sig figs only, The zeros in this number are not sig figs as they are just holding the place values. As the least number of sig figs here is two, the answer needs to be reported with two sig figs only.

\frac{4.03}{0.0000035}=1.2*10^6



4 0
2 years ago
Identify the MOs that react to form cyclohexene. HOMO of 1,3-butadiene and LUMO of ethylene LUMO of 1,3-butadiene and LUMO of et
BARSIC [14]

Answer:

HOMO of 1,3-butadiene and LUMO of ethylene

HOMO of ethylene LUMO of 1,3-butadiene

Explanation:

1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.

According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.

6 0
2 years ago
Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30
FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

10^{-1.30} = \frac{x^2}{0.09-x}

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

<em>pH = 0,0301</em>

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 1,30</em>

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

10^{-6.70} = \frac{x^2}{0.09-x}

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

<em>pH = 2,87</em>

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 6,70</em>

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

5.01x10^{-8} = \frac{x^2}{0.09-x}

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

<em>pH = 10,7</em>

<em></em>

I hope it helps!

6 0
2 years ago
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