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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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Which of the following electron configurations are written incorrectly?
Lynna [10]

Answer:

The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.

Explanation:

The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.

There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.

All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".

In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.

The other options are correctly written.

3 0
3 years ago
How much water, in grams, can be made from 1.03 x <img src="https://tex.z-dn.net/?f=10%5E%7B24%7D" id="TexFormula1" title="10^{2
krok68 [10]

Answer:

30.8 g of water are produced

Explanation:

First of all we need the equation for the production of water:

2H₂ + O₂ → 2H₂O

2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.

As we assume, the oxygen in excess, we determine the moles of H₂.

1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles

Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water

Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced

5 0
3 years ago
WILL GIVE BRAINLIEST AND 5 STARS what is the purpose/ use of a Bunsen Burner
Roman55 [17]

A Bunsen burner, named after Robert Bunsen, is a common piece of laboratory equipment that produces a single open gas flame, which is used for heating, sterilization, and combustion. The gas can be natural gas (which is mainly methane) or a liquefied petroleum gas, such as propane, butane, or a mixture of both. Have A Great Day :)

5 0
3 years ago
Read 2 more answers
Any two difference between short sightedness and long sightednes​
SSSSS [86.1K]

Explanation:

tala it is also called myopia lekhnu hai

Äni Arko MA it is also called hypermetropia .

3 0
2 years ago
3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa
Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

7 0
3 years ago
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