All categories

4 years ago

What is necessary for the transport of oxygen by an erythrocyte?

1 answer:

6 0

I think the correct answer is Hemoglobin. Hemoglobin is necessary for the transport of oxygen by an erythrocyte. It is a protein molecule found in red blood cells made of four subunits; two alpha subunits and two beta subunits. This protein is responsible for carrying of oxygen to body cells and tissues, and carbon dioxide to the lungs.

You might be interested in

You are given a small bar of an unknown metal. You find the density of the metal to be 10.5 g/cm3. An X-ray diffraction experime

Artemon [7]

Answer:

Explanation:

5 0

4 years ago

Read 2 more answers

What does it mean when a wave's amplitude increases?

anastassius [24]

Answer:

It becomes shorter

Explanation:

A wave, such as a Soundwave, controls volume based on it's Wavelength. A Louder sound has a shorter wavelength, a softer sound has it's waves spread out more. Amplitude is simply a fancy name for Volume in Amps. When amplitude increases, the volume increases. Thus, waves become shorter.

7 0

4 years ago

Read 2 more answers

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70

kakasveta [241]

Explanation:

The given data is as follows.

$P_{atm}$ = 98.70 kPa = 98700 Pa,

T = $30^{o}C$ = (30 + 273) K = 303 K

height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = $13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}$

= 13534 $kg/m^{3}$

The relation between pressure and atmospheric pressure is as follows.

P = $P_{atm} + \rho gh$

Putting the given values into the above formula as follows.

P = $P_{atm} + \rho gh$

= $98700 Pa + 13534 \times 9.81 \times 0.03 m$

= 102683.05 Pa

= 102.68 kPa

thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.

8 0

4 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Len [333]

Answer:

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²) where λ is the wavelength

Rh is Rydberg constant

n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a. n₁ = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b. n₁ = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/4² - 1/5²) = 2.468 x 10⁵/m

λ = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x (1 x10⁹ nm/m) = 4051 nm

c. n₁ = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/5²) = 7801 x 10⁵/m

λ = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

8 0

3 years ago