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den301095 [7]
3 years ago
7

How do cells reproduce? ∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵

Chemistry
1 answer:
lora16 [44]3 years ago
8 0

Answer:

Cells divide to reproduce. There are two primary methods used, one for somatic cells, which compose the organism’s body, and one for reproductive cells, or gametes.

Scientists call the process of somatic cell division mitosis. Mitosis has six distinct steps in which the cell organizes and copies the DNA in the nucleus. Once copied, each new cell has its own copy of the DNA. The six steps of mitosis are prophase, prometaphase, metaphase, anaphase, telophase and cytokinesis. Some authorities consider the non-dividing portion of the cell’s lifecycle, known as interphase, to be one of seven steps involved in mitosis; however, interphase cells are not actively dividing.

Sex cells, such as sperm or eggs, must divide differently. Sex cells only possess one-half of the DNA that makes up a new animal. That way, when they combine, the resulting organism gets half of its DNA from its mother and half from its father. Because sex cells only want half of the DNA in each cell, they go through a different division process called meiosis. In meiosis, the cells split a second time, which yields four daughter cells rather than two as with mitosis; however, this provides each daughter cell with 23 chromosomes in contrast to the 46 chromosomes in somatic cells.

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Convert 13.4 degrees celcius into kelvin​
Alenkasestr [34]

Answer:

286.55K

Explanation:

To convert to kelvin , add 237 .15

13.4\°C + 273.15 = 286.55K

5 0
3 years ago
Read 2 more answers
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. Based on the r
SashulF [63]

Answer:

Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.

However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.

3 0
3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

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

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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Ilia_Sergeevich [38]

Answer:

oil

Explanation:

i hope this helps :)

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