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3 years ago

In tissue engineering, scaffolds are built for growing replacement body parts primarily so that:

2 answers:

7 0

One of the functions of scaffolds in tissue engineering is letter b. cells grow into the correct shape. Scaffolds used in tissue engineering have biomaterials with biological attributes <span>that will enhance cell attachment </span>and topography where cells morph and align.

victus00 [196]3 years ago

5 0

Answer: Option B

Explanation: In case of tissue engineering, scaffolds are built for growing replacement body parts. This is done to provide a proper shape for the growth of the cells.

The cells in the body needs to grow in a proper shape and size means it should in the affected area only.

This will result in on time growth and development of the affected area as scaffold will provide a base for the growth of cells in proper shape.

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Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

Explanation :

The given two chemical reactions are:

(1) $2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

(2) $3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) $6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)$

Now we are adding both the reactions, we get the overall chemical reaction.

$6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

The $MnO_2$ is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

5 0

3 years ago

Water and oxygen gas are the products of a chemical reaction.

Darina [25.2K]

The answer is C.

H₂O₂ ----> H₂O + O₂

8 0

3 years ago

Read 2 more answers

.22

Alja [10]

Answer:its b

Explanation:

7 0

3 years ago

Read 2 more answers

The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond

kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

ΔGº= - RT ln Ka

ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

6 0

3 years ago

Disadvantages of Mendeleev's periodic table

mestny [16]

Answer:

Various limitations of Mendeleev's periodic table are:-

Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .

Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.

In the d-block , elements with lower atomic number were placed before higher atomic number.

Explanation:

3 0

3 years ago