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vova2212 [387]
4 years ago
7

A red blood cell has a round structure. A muscle cell has a long, thin structure. Which statement best explains why these two ty

pes of cells have different structures
Chemistry
1 answer:
Sphinxa [80]4 years ago
3 0

Answer:

The red balloons have a biconcave disk structure, which is what is worded in the question seems to me a little wrong.

On the other hand, the elongated structure of muscle cells is due to a greater initial length in order to perform muscle contractions in a better way, or to perform their function at sea.

Something very important that the differences is that the red balloon does not have a nucleus (the muscle cell is multinucleated and has more mitochondrial load) and this structure is characteristic for being able to transport gases such as oxygen and disseminate them better in the body.

Explanation:

Red blood cells also have a shorter half-life than muscle cells, they are responsible for traveling throughout the body and renew their hemoglobin component after 120 days in the spleen.

On the other hand it is important to note that both work together when performing physical exercise, one donates oxygen and the other consumes it, considering that the muscle cell has more capacity to withstand hypoxic moments, that is why it supports lactic fermentations in anaerobic processes .

In summary, the red globule has this shape to act as gas transport, and the muscle cell to have a better start in contractile function.

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Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init
Jlenok [28]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

                          H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
4 years ago
How many grams of naoh would react with 507 g fecl2 in the reaction fecl2 + 2naoh fe(oh)2(s) + 2nacl?
rosijanka [135]
<span>Answer: Correct answer is 507g FeCl2 x (1 mol FeCl2 / 126.8 g FeC2) x (1 mol Fe(OH)2 / 1 mol FeCl2) x (89.8 g Fe(OH)2/ 1 mol Fe(OH)2) = 359 g Fe(OH)2.</span>
7 0
3 years ago
Read 2 more answers
Determine the quality of a two-phase liquid-vapor mixture of (a) H.O at 100°C with a specific volume of 0.8 m /kg. (b) Refrigera
Paul [167]

Answer:

a) 0.478 4

b) 1.45 × 10⁻⁶

Explanation:

A two phase liquid-vapor mixture

a)  Given:

Temperature of H.O = 100°C

Specific volume of mixture, v = 0.8

now, we know

The specific volume of water at 100°C, vf = 0.001043 m³/kg

and,specific volume of steam At 100°C , vg = 1.671 m³/kg

also,

Specific volume, v = [ vf + x × (vg - vf )  ]

here, x is the quality

therefore, we get

0.8 = 0.001043 + [ x × (1.671 - 0.001043) ]

or

0.798957 = x × 1.669957

or

x = 0.478 4

b)  Temperature of refrigerant = 0°C

Specific volume of refrigerant, v = 0.7721 cm³/g = 0.7721 × 10⁻³ m³/kg

Specific volume of fluid, vf = 7.72 × 10⁻⁴ m³/kg

Specific volume of gas, vg = 0.06930 m³/kg

Now,

Specific volume of two-phase mixture, v = [ vf + x × (vg - vf )  ]

on substituting the values in the above equation, we get

0.7721 × 10⁻³ = 7.72 × 10⁻⁴ + [ x × (0.06930 - 7.72 × 10⁻⁴ )  ]

or

x = 1.45 × 10⁻⁶

4 0
3 years ago
A potato gun beautifully illustrates Law. As you decrease the volume of the air in between two pieces of potato, the
kati45 [8]

<u>Answer:</u> The law that illustrates this is Boyle's Law

<u>Explanation:</u>

  • Boyle's law: This law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

Mathematically,

P\propto \frac{1}{V}         (at constant temperature and number of moles)

  • Charles' law: This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

V\propto T         (at constant pressure and number of moles)

  • Avogadro's law: This law states that volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.

Mathematically,

V\propto n         (at constant pressure and temperature)

So, when the volume of the air in the gun decreases, the pressure is increased. The inverse relationship between the volume and pressure of the gas is very well illustrated by Boyle's Law.

Hence, the law that illustrates this is Boyle's Law

8 0
4 years ago
An igneous rock originally has 3 grams of uranium 238 in it. When dated the rock only contains 1.8 grams. What are the parent an
castortr0y [4]

Answer :

The parent and daughter concentrations (in percentages) is, 60 % and 40 % respectively.

The age of rock is 3.32\times 10^9\text{ years}

Explanation :

First we have to calculate the parent and daughter concentrations (in percentages).

\text{Parent concentrations}=\frac{1.8g}{3g}\times 100=60\%

and,

\text{Daughter concentrations}=\frac{(3-1.8)g}{3g}\times 100=40\%

As we know that, the half-life of uranium-238 = 4.5\times 10^9 years

Now we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{4.5\times 10^9\text{ years}}

k=1.54\times 10^{-10}\text{ years}^{-1}

Now we have to calculate the time passed.

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  = 1.54\times 10^{-10}\text{ years}^{-1}

t = time passed by the sample  = ?

a = initial amount of the reactant  = 3 g

a - x = amount left after decay process = 1.8 g

Now put all the given values in above equation, we get

t=\frac{2.303}{1.54\times 10^{-10}}\log\frac{3}{1.8}

t=3.32\times 10^9\text{ years}

Therefore, the age of rock is 3.32\times 10^9\text{ years}

5 0
3 years ago
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