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2 years ago

The production of new skin cells to replace those that have been lost is an example of what characteristic of life?

Chemistry

1 answer:

Alchen [17]2 years ago

5 0

The production of new skin cells is example of regeneration.

Regeneration is process of replacing or restoring not just skin cells, but also other cells in the human organism.

Skin regeneration is replacement of damaged tissue with new tissue.

There are two ways of skin regeneration:

1) reconstruction is a process of rebuilding of damaged skin cells

2) restoration is process of replacing broken cell skins

Epidermal stem cells are those who produces new daughter cell skins.

Epidermal stem cells are in the lowest layer of the skin.

More about skin: brainly.com/question/306377

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You might be interested in

Help I don’t know what the answer is

irina [24]

Answer:-

10.0

Explanation: -

The equation is

N2O4 = 2NO2.

Concentration of NO2 = [NO2} = 0.500 M

Concentration of N2O4 = [N2O4] = 0.0250 M

Keq is the equilibrium constant.

It is the ratio of the concentration of products raised to the stoichiometric coefficient of the product to the concentration of the reactants raised to the stoichiometric coefficient of the reactants.

The expression for Keq = [NO2] ^2 / [N2O4]

= 0.500^2 / 0.0250

= 10.0

6 0

3 years ago

Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

$HONH_2+HCl\rightarrow HONH_3^+Cl^-$

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

$V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL$

Now, we can see that the moles of acid, base and acidic salt are all:

$0.0220L*0.200mol/L=0.0044mol$

And therefore the concentration of the salt at the equivalence point is:

$[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M$

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

$HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2$

Whereas the equilibrium expression is:

$Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}$

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}} \\\\Ka=1.10x10^{-6}$

So we can write:

$1.10x10^{-6}=\frac{x^2}{0.0858-x}$

And neglect the x on bottom to obtain:

$1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M$

And since x=[H3O+] we obtain the following pH:

$pH=-log(3.07x10^{-4})\\\\pH=3.513$

Regards!

4 0

3 years ago

Suppose a vodka martini contains 30% alcohol with the remaining portion of the drink composed of water. what is the solute in th

viktelen [127]

Solute - substance that dissolves.
alcohol - solute
water - solvent

7 0

4 years ago

The questions are attached

Sladkaya [172]

Answer:

Explanation:

a) Balanced chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

b) Mass of oxygen gas needed = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/ 44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of oxygen and propane:

C₃H₈ : O₂

1 : 5

0.57 : 5/1×0.57 = 2.85 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 2.85 mol × 32 g/mol

Mass = 91.2 g

c) water vapors produced = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/ 44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of water and propane:

C₃H₈ : H₂O

1 : 4

0.57 : 4/1×0.57 = 2.28 mol

Mass of water vapors:

Mass = moles × molar mass

Mass = 2.28 mol × 18 g/mol

Mass = 41.0 g

mass of water vapors produced = 1.5 kg (1500 g)

Mass of propane reacted = ?

Solution:

Number of moles of water vapors:

Number of moles = mass/molar mass

Number of moles = 1500 g/ 18 g/mol

Number of moles = 83.33 mol

now we will compare the moles of water and propane.

H₂O : C₃H₈

4 : 1

83.3 : 1/4×83.3 = 20.8 mol

Mass of propane:

Mass = number of moles ×molar mass

Mass = 20.8 mol × 44.1 g/mol

Mass = 917 g

3 0

3 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

Learn more about pH here:

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8 0

2 years ago