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SSSSS [86.1K]
3 years ago
10

How does the process of cell division replace damaged cells

Chemistry
1 answer:
ollegr [7]3 years ago
3 0

Answer:

A

Explanation:

The cells split up into two to repair the damaged cell

You might be interested in
How many moles are in 2.4g of carbon dioxide (CO2)?
miv72 [106K]

Answer:

\boxed {\boxed {\sf 0.055 \ mol \ CO_2}}

Explanation:

To convert form grams to moles, the molar mass must be used. This is the mass (in grams) in 1 mole of a substance.

We can use the values on the Periodic Table. First, find the molar masses of the individual elements: carbon and oxygen.

  • C: 12.011 g/mol
  • O: 15.999 g/mol

Check for subscripts. The subscript of 2 after O means there are 2 oxygen atoms, so we have to multiply oxygen's molar mass by 2 before adding.

  • O₂: 2* (15.999 g/mol)=31.998 g/mol
  • CO₂: 12.011 g/mol + 31.998 g/mol =40.009 g/mol

Use the molar mass as a ratio.

\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}

Multiply by the given number of grams.

2.4 \ g \ CO_2 *\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}

Flip the fraction so the grams of carbon dioxide cancel.

2.4 \ g \ CO_2 *\frac { 1 \ mol \ CO_2}{44.009 \ g\ CO_2}

2.4  *\frac { 1 \ mol \ CO_2}{44.009}

\frac { 2.4 \ mol \ CO_2}{44.009}= 0.0545342998 \ mol \ CO_2

The original measurement of grams has 2 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

The ten thousandth place has a 5, so we round the 4 to a 5.

0.055 \ mol \ CO_2

2.4 grams of carbon dioxide is about 0.055 moles.

8 0
3 years ago
If you add 90 mL of 0.15M stock solution, to 10 mL of water, what is the new concentration?
lyudmila [28]

0,15 moles of NaOH-------in------------1000ml

x moles of NaOH------------in--------100ml

x = 0,015 moles of NaOH

final volume = 150ml

0,015 moles of NaOH---in-------150ml

x moles of NaOH--------------in-----1000ml

x = 0,1 moles of NaOH

answer:  0,1mol/dm³  (molarity)

6 0
3 years ago
how does the total mass of each object increases the amount of force that is needed to get them moving at 5 m/s increase by abou
irakobra [83]

Answer:

The answer is below

Explanation:

Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum with respect to time, going in the same direction as the force.

Let F = force, m = mass of object, v = velocity of object, mv = momentum.

F = d/dt(mv) = m(dv / dt) = ma; a = acceleration.

Let us assume that the object starts from rest to 5 m/s within 1 seconds, hence:

F = m(dv / dt)

200 N = m[(5 m/s - 0 m/s) / (1 s)]

200 = 5m

m = 40 kg

7 0
3 years ago
What’s the correct answer
Fed [463]
The correct answer is: 
_________________________________________________________
    " F  and Br , because they are in the same group" .
_________________________________________________________
Note:
_________________________________________________________
Choice [B]:  "F and Br ; because they are in the same period" ; is incorrect; since  "F" and "Br" are not in the same "period" (that is, "row").
______________________________________________________
Choice [C]:  "Na and Mg; because they are in the same group {"column"} ;  is incorrect; since:  "Na" and "Mg are NOT in the same group {"column"].
_______________________________________________________
Choice [D]:  "Na and Mg" ; because they are in the same period {"row"}; is incorrect.  Note:  "Na" and "Mg" are, in fact,  in the same period {"row"}. However, as aforementioned, {Mg and Na} are not in the same group {"column".}.  

Note:  The similiarities in physical and chemistry properties among elements are determined and organized — or tend to be so—by "groups" {"columns"} — NOT by "periods" {rows}.  
______________________________________________________
3 0
4 years ago
Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a
nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

A->B;\Delta _rG^o=-13 kJ/mol

C ->D ;\Delta _rG^o=3.5 kJ/mol

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

6 0
3 years ago
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