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tekilochka [14]
3 years ago
9

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:
SIZIF [17.4K]3 years ago
5 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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kkurt [141]

a) 56g

<h3>Calculation:</h3>

At STP,

22.4 L of N₂ = 1 mol

We have given 44.8 L of N₂, therefore,

44.8 L of N₂ = \frac{44.8}{22.4}

                    = 2 mol

We know that,

1 mol of N₂ = 28 g

Hence,

2 mol of N₂ = 28 × 2

                   = 56g

Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.

Learn more about calculation at STP here:

brainly.com/question/9509278

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3 0
2 years ago
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If 0.8675 g of KHP requires 24.56 mL of an NaOH solution to reach the equivalence
ValentinkaMS [17]

Answer:

M_{base}=0.173M

Explanation:

Hello there!

In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:

n_{acid}=M_{base}V_{base}

Thus, by solving for the volume of the base as NaOH, we obtain:

M_{base}=\frac{n_{acid}}{V_{base}} \\\\M_{base}=\frac{0.8675g*\frac{1mol}{204.22g} }{0.02456L} \\\\M_{base}=0.173M

Best regards!

6 0
3 years ago
Someone please help! this is the last question<br>I only need help with B.<br><br>​
ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

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2 years ago
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kirill [66]

Answer:

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Sugar is the solute since it is the thing that is being dissolved.
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