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jenyasd209 [6]
2 years ago
5

our throat is dry, and you want the last cough drop in the box to last a long time in your mouth. What should you do? Keep the c

ough drop whole. This maintains the largest surface-to-volume ratio and slows the dissolution of the cough drop. Break the cough drop into little pieces and put them all in your mouth. Since each little piece must be dissolved separately, the drop will last longer. It doesn't matter if the cough drop is in one piece or many pieces; the total amount of cough drop is all that matters. Break the cough drop into little pieces and put them all in your mouth. This decreases the surface-to-volume ratio and slows the dissolution of the cough drop.
Chemistry
1 answer:
Ann [662]2 years ago
8 0

You should keep the cough drop whole. This maintains the largest surface-to-volume ratio and slows the dissolution of the cough drop.

<h3>Delay of drug action</h3>

The action of the drug can be delayed to either reduce adverse effect if the drugs to the body or to enhance its therapeutic purposes.

Because the throat is dry, the therapeutic purposes of the cough drop would only be achieved if it's dissolution is delayed in the mouth.

To do so, the cough drop should be kept whole to maintain the largest surface-to-volume ratio. This would slow the dissolution of the cough drop.

Learn more about drug here:

brainly.com/question/26254731

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20.00 g of aluminum (al) reacts with 78.78 grams of molecular chlorine (cl2) all of each reactant is completely consumed and a s
marysya [2.9K]

The mass of the product is <em>98.78 g.</em>

The word equation is

aluminum + chlorine → product

20.00 g + 98.78 g → <em>x</em> g

If each reactant is completely consumed, the <em>Law of conservation of Mass </em>tells us the mass of the product must be 98.78 g.

8 0
3 years ago
Please help quick
Rama09 [41]

Answer:

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g ×  × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g ×  c ×  -64.8°C

7020 J = -7821.36 g.°C ×  c

c = 7020 J / -7821.36 g.°C

c = 0.898 J/g.°C

Negative sign shows heat is released.

7 0
3 years ago
I was checking my answers for a chem test and came across a solution for a problem where the ∆S was pos, ∆H was neg and ∆G was n
nevsk [136]

Answer:

there is only 15 points

Explanation:

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: \Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0ΔS  

universe

​

=ΔS  

system

​

+ΔS  

surroundings

​

>0delta, start text, S, end text, start subscript, start text, u, n, i, v, e, r, s, e, end text, end subscript, equals, delta, start text, S, end text, start subscript, start text, s, y, s, t, e, m, end text, end subscript, plus, delta, start text, S, end text, start subscript, start text, s, u, r, r, o, u, n, d, i, n, g, s, end text, end subscript, is greater than, 0

At constant temperature and pressure, the change in Gibbs free energy is defined as \Delta \text G = \Delta \text H - \text{T}\Delta \text SΔG=ΔH−TΔSdelta, start text, G, end text, equals, delta, start text, H, end text, minus, start text, T, end text, delta, start text, S, end text.

When \Delta \text GΔGdelta, start text, G, end text is negative, a process will proceed spontaneously and is referred to as exergonic.

The spontaneity of a process can depend on the temperature.

Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:

8 0
3 years ago
Try to make only one molecule of water in the simulation. Is it possible?
mel-nik [20]

Answer:

A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.

Explanation:

3 0
3 years ago
How is light produced in an atom?
Nadya [2.5K]

Answer:

light is the result of electrons moving between defined energy levels in an atom called shells.

Explanation:

when something exited an atom like collision with another atom or a chemical reaction, an electron may absorb energy boosting it to a higher level shell.

3 0
3 years ago
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