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3 years ago

What is the definition of cleavage

Chemistry

1 answer:

aev [14]3 years ago

7 0

A division or split

in biology a cell division

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Which of the following is the quantum number set for an electron in the 3rd energy level, dumb-bell shaped orbital, on the z-axi

Juli2301 [7.4K]

Answer:

Explanation:

Principal quantum no "n" = 3

Azimuthal quantum no "l"= 1

Magnetic quantum no "m"= +1/2

Over all is 3pz

5 0

3 years ago

Oxygen always has an oxidation number of -2 unless it is combined with fluorine or found in the compound peroxide. When in the c

Tpy6a [65]

When oxygen is found is peroxide, it has an oxidation number of -1.

The chemical formula of hydrogen peroxide is H2O2. We know that hydrogen always has +1 oxidation state until it forms metal hydrides. So in H2O2, the oxidation state ofhydrogen is +1.

Now, let oxidation state of oxygen be x. So,

2 * (+1) + 2*x = 0

2 + 2x = 0

2x = -2

x = -2 / 2

x = -1

Hence, the oxidation number of oxygen in peroxides is -1

6 0

3 years ago

A recent study has revealed that chlorinated hydrocarbons, gasoline and other volatile organic compounds (VOC's) have become sig

Sedaia [141]

Answer:

ground water / leaking storage tanks

5 0

3 years ago

Can magnesium sulphate be used for water coagulation

sweet [91]

Yes it can cause you put epsom salt in a bath and theirs magnesium in that and it’s good for your body and takes fat from your body

3 0

3 years ago

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

Sholpan [36]

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

4 0

3 years ago