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maria [59]
3 years ago
7

Which explains why the ratio of cell surface area to volume affect the cell size? PLEASE HELP ASAP! THIS IS DUE TOMORROW!

Chemistry
1 answer:
diamong [38]3 years ago
5 0
Think about it this way: a cell can only take in materials through the cell membrane. Naturally, as the cell membrane surface area increases, then the amount of material that can enter the cell increases due to more entry points along the membrane. However, when the cell increases in size then the volume inside the cell will also increase - more volume inside requires more energy to transport materials around the cell. So, there is a trade-off between the surface area and volume. 
<span>As an example, consider a sphere as a cellular model. The surface area of a sphere is </span>

<span>SA = 4*pi*r^2 </span>

<span>while the volume of the sphere is </span>

<span>V = 4/3*pi*r^3 </span>

<span>initially, as a very small cell increases in radius, the surface area will increase at a greater rate than the volume. But as the cell gets bigger there will be a point where the volume increases faster than the surface area. Cells have maximized this ratio through evolution (this is also one reason why we are not single-celled organisms). </span>
<span>Some cells are able to get around this issue to some extent by "folding" the membrane, thus increasing the surface area without affecting the volume by much. </span>
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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
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Mamont248 [21]

Answer:

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6 0
3 years ago
What element has 2 electrons in its outermost energy level and is in the 2nd period on periodic table?
Olegator [25]
Be-beryllium have 2 electrons and it is in the 2 nd period
4 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
Calculate the pH of a solution created by placing 2.0 grams of yttrium hydroxide, Y(OH)3, in 2.0 L of H2O. Ksp for Y(OH)3 is 6.0
oee [108]

Answer:

pH = 8.314

Explanation:

  • Y(OH)3(s) ↔ Y+  +  3OH-

equil:   S               S         3S

∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24

⇒ 6.0 E-24 = ( S )*( 3S )³

⇒ 6.0 E-24 = 27S∧4

⇒ 2.22 E-25 = S∧4

⇒ ( 2.22 E-25 )∧(1/4) = S

⇒ S = 6.866 E-7 M

⇒ [ OH- ] = 3*S =2.06 E-6 M

⇒ pOH = - Log [ OH- ]

⇒ pOH = - Log ( 2.06 E-6 )

⇒ pOH = 5.686

∴ pH = 14 - pOH

⇒ pH = 8.314

8 0
3 years ago
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