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vladimir2022 [97]

2 years ago

8

What are the two chemical reactions that are responsible for making cooked food so tasty?

2 answers:

KengaRu [80]2 years ago

8 0

Nose what .,...,).),),,

Eva8 [605]2 years ago

8 0

Answer:

Mailard process produces flavour and aroma during the cooking process, and it is used almost everywhere from the baking industry to our day to day life to make food tasty it is also called nonenzymatic browning reaction since it takes place in the absence of enzymes.

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1. which is not a component of biodiversity

Vera_Pavlovna [14]

The answer is C, i just took the k-12 test~

8 0

2 years ago

The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

3 0

2 years ago

Please hurry this is due tomorrow morning

Llana [10]

Answer: G Atom 1 and Atom 4.. hope this helps and good luck!

3 0

2 years ago

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago

What is the volume of one mole of any gas at STP?

Vikki [24]

The volume of one mole of any gas at Standard Temperature and Pressure (1 atm and 0 degrees Celsius [273K]) is 22.4 L.

7 0

3 years ago