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Svetlanka [38]
1 year ago
14

Describe ways a lack of water could affect cell functions?

Chemistry
1 answer:
ladessa [460]1 year ago
4 0

The way a lack of water could affect cell functions is that cell membranes would lack structure.

Osmosis causes water to flow from the cell into the surrounding liquid. This causes the protoplasm all the material inside the cell, to shrink from the cell wall. Excessive water loss leading to cell wall breakdown can lead to cell death. Water is an important part of the protoplasm of living cells as it is directly involved in myriad biochemical reactions such as photosynthesis and respiration.

Without them, cells cannot move waste and byproducts absorb nutrients, carry out intracellular transport functions or send signals. Regulates body temperature. Moisten eye nose and mouth tissues. Protects body organs and tissues. Carry nutrients and oxygen to cells. Plasmolysis is a situation in which cells lose water in hypertonic solutions. the plant cells lose water causing them to shrink or shrink removing the cell cytoplasm from the cell wall.

Learn more about Cell functions here:-brainly.com/question/13880600

#SPJ9

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Answer:

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Explanation:

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3 years ago
Which of the following waves have the shortest wavelengths?
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This goes from longest to shortest wavelengths
Radio waves, microwaves, infrared, light, ultra violet, X-rays, gamma rays
4 0
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Read 2 more answers
Explain how the copper could be in the lake sample near the picnic area but not have been detected by this test.
Masja [62]

Answer:

May be the instrument is incorrect or may be error in it.

Explanation:

The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one  or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.

8 0
3 years ago
(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t
11Alexandr11 [23.1K]

Answer:

a

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

b

The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

(1)

The position of an atom that fits in the octahedral site with radius \left( r \right)is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

2r=a-2R------(A)

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider \Delta {\rm{XYZ}} as follows:

(XY)^ 2 =(YZ) ^2 +(XZ)^2

Substitute XY as{\rm{R}} + 2{\rm{R + R}} and {\rm{YZ}} as a and {\rm{ZX}} as a in above equation as follows:

(R+2R+R) ^2 =a ^2 +a^ 2\\16R ^2 =2a^ 2\\ a =2\sqrt{2R}

Substitute value of aa in equation (A) as follows:

r= \frac{2\sqrt{2}R -2R }{2} \\ =\sqrt{2} -1R\\ = 0.414R

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.  

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

AD=2R+2r----(B)

   In    \Delta {\rm{ABC}},

(AB) ^2 =(AC) ^2 +(BC) ^2

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute {\rm{AC}} as a and {\rm{BC}} as a in above equation as follows:

(AB) ^2 =a ^2 +a ^2

AB= \sqrt{2a} ----(1)

Also,

AB=2R

Substitute value of 2{\rm{R}} for {\rm{AB}} in equation (1) as follows:

2R= \sqrt{2} aa = \sqrt{2} R

Therefore, the length of body diagonal is calculated using Pythagoras Theorem in \Delta {\rm{ABD}} as follows:

(AD) ^2 =(AB) ^2 +(BD)^2

Substitute {\rm{AB}} as \sqrt 2a   and {\rm{BD}} as a in above equation as follows:

(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

\sqrt 3a=2R+2r

Substitute a as \sqrt 2{\rm{R}} in above equation as follows:

( \sqrt3 )( \sqrt2 )R=2R+2r\\\\

r = \frac{2.4494R-2R}{2}\\

=0.2247R

\approx 0.225R

The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Note

An impure atom occupies the tetrahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The length of body diagonal is calculated using Pythagoras Theorem. The body diagonal is equal to the sum of the radii of two atoms. This helps in determining the relation between the radius of impure atom and radius of atom present in the unit cell.

7 0
3 years ago
An electrochemical cell has the following standard cell notation.
Alona [7]

Explanation:

The given following standard cell notation.  

Mg(s) | Mg^2+ (aq) || Aq^+(aq) | Aq(s)

Oxidation:

Mg(s)\rightarrow Mg^{2+}+2e^-....(1)

Magnesium metal by loosing 2 electrons is getting converted into magnesium cation. Hence, getting oxidized

Reduction:

Ag^+(aq)+1e^-\rightarrow Ag(s)...(2)

Silver ion by gaining 1 electrons is getting converted into silver metal. Hence, getting reduced.

Overall redox reaction: (1)+2 × (2)

Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}+2Ag(s)

4 0
3 years ago
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