answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul
Free_Kalibri [48]
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4
8 0
1 year ago
Read 2 more answers
Calculate the ratio of effusion rates of cl2 to f2 .
Lelechka [254]
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
4 0
2 years ago
If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
Arada [10]
Find moles of MgSO4.7H2O

molar mass = 246
so moles  = 32 / 246  = 0.13 moles.

When heated, all 7 H2O from 1 molecule will be gone. 
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18  = 16.38g

so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g
6 0
1 year ago
The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g
valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

5 0
1 year ago
A student hears in chemistry class that a certain object is made of a pure substance. What can the student conclude about the ob
umka21 [38]
It has no other elements
5 0
1 year ago
Read 2 more answers
Other questions:
  • Which equations shows the complete dissociation of a strong base?
    9·2 answers
  • The Kp for the reaction below is 1.49 × 108 at 100.0°C:
    5·2 answers
  • Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
    7·1 answer
  • At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.
    8·2 answers
  • In the manufacture of steel, pure oxygen is blown through molten iron to remove some of the carbon impurity. if the combustion o
    13·2 answers
  • Hydrogen sulfide decomposes according to the following reaction, 2 H2S(g) → 2 H2(g) + S2(g) where ∆S = +78.1 J/K, ∆H = +169.4 kJ
    13·1 answer
  • one kilogram of water (V1 = 1003 cm^3*kg-1) in a piston cylinder device at (25°C) and 1 bar is compressed in a mechanically reve
    6·1 answer
  • (2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
    13·1 answer
  • What evidence did the team examine that matter is conserved when dry ice changes into a gas?
    10·1 answer
  • Lana is testing her hypothesis that marigolds grow faster in red light than in yellow light. If the plants in yellow light grow
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!