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1 year ago

Which of the following correctly describes the process of inspiration (air entering the lungs).

Chemistry

1 answer:

IgorC [24]1 year ago

3 0

The lungs expand, causing their internal pressure to decrease.

Explanation:

Inhalation is one of the two mechanisms of breathing when air or oxygen is taken in.

During the process of inhalation or inspiration when air is inhaled following changes occur:

Diaphragm gets contracted and pulls to downward attaining convex shape and the muscles in between rib ( intercostal muscles) get contracted and gets upward pull attaining the concave shape.

This contraction causes the lungs to expand and the air pressure with in gets decreased in comparison to atmospheric pressure. This results in inhalation of air into the lungs.

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Americium-242 has a half-life of 6 hours. If you started with 24 g and you now have 3 g, how much time

kogti [31]

How many times has it halved?

24/2 = 12
12/2 = 6
6/2 = 3

It halved three times.
It halves once every 6 hours.

18 hours have passed.

7 0

2 years ago

One atom of silicon can properly combined in a compound with

xxMikexx [17]

The correct answer should be two oxygen atoms. That's because it's properties are similar to carbon insofar that it can form four bonds, so if it forms bonds with 2 oxygen atoms then it will have all four bonds created since Oxygen forms double bonds. This would make SiO2 which is also known worldwide as silica.

8 0

2 years ago

A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what

FrozenT [24]

Answer:

C. 481 °C.

Explanation:

At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

To find the amount of heat, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

∵ Q of copper = Q of water

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

So, the right choice is: C. 481 °C.

4 0

2 years ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

What is the molarity of a solution made by dissolving 8.60 g of a solid with a

Dima020 [189]

Answer:

1.43 M

Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

Mass of solid = 8.60 g

Molar mass of solid = 21.50 g/mol

Mole of solid =?

Mole = mass / molar mass

Mole of solid = 8.60 / 21.50

Mole of solid = 0.4 mole

Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

280 mL = 280 mL × 1 L / 1000 mL

280 mL = 0.28 L

Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

Mole of solid = 0.4 mole

Volume = 0.28 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.4 / 0.28

Molarity = 1.43 M

Thus, the molarity of the solution is 1.43 M.

8 0

2 years ago