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harina [27]
2 years ago
10

1. Draw the basic repeating unit of sodium alginate polymer. 2. What is this polymer used for?

Chemistry
2 answers:
tamaranim1 [39]2 years ago
8 0

Answer:

1.The diagram is attached

2.sodium alginate is a polymer which can be extracted from brown seaweed and is used for defloculating, gelling and thickening

Used also in dermatological preparation

Used in adhensive paste

Blizzard [7]2 years ago
8 0

Answer:

The structure is shown in the attached file.

The Molecular Formula of sodium alginate is (C6H7NaO6)n or C6H9NaO7

Uses of sodium alginate polymer

1. Boiler Water Additive

2. Food Additives: such as emulsifier, gelling_agent, stabilizer, thickener, et.c.

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Suppose that a certain fortunate person has a net worth of $79.0 billion ($7.90×1010). If her stock has a good year and gains $3
mario62 [17]
A net worth: $79.0 billion.
Value of stock : $3.20 billion.
New net worth:
$79.0 + $3.20 = $82.20 billion = $82.20 * 10^9 = $8.20 * 10^10
7 0
2 years ago
Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34 × 10 − 59 2.34×10−59 , and d
Margarita [4]

Answer:

1.315x10⁻³M = [Ca²⁺]

Explanation:

Based in the reaction:

Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)

Solubility product, ksp, is defined as:

ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²

From 1 mole of hydroxyapatite are produced  10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:

6/10 Ca²⁺ = PO₄³⁻

Replacing in ksp formula:

ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²

As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:

2.34x10⁻⁵⁹ =  [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²

3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶

<em />

<em>1.315x10⁻³M = [Ca²⁺]</em>

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I hope it helps!

5 0
2 years ago
The terms motif (fold) and domain describe levels of protein organization more complicated than primary or secondary structure.
posledela

Answer:

Each specific property of motif and domain is explained.

Explanation:

Domain;

  • May retain a 3D structure when separated from rest of the protein.          
  • Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
  • Stable globular units like pyruvate kinase
  • May be distinct functional units in a protein

Motif;

  • Repetetive supersecondary structure because they contain cluster of secondary structure.
  • Beta Alpha Beta unit is an example of motif
  • Clusters of secondary structure

Both Motif and Domain;

  • Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
  • Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.

8 0
2 years ago
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S
Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of ^{235}UF_6=349.034348g/mol

Molar mass of ^{238}UF_6=352.041206g/mol

By taking their ratio, we get:

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}

From the above relation, it is clear that rate of effusion of ^{235}UF_6 is faster than ^{238}UF_6

Difference in the rate of both the gases, Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816

To calculate the percentage increase in the rate, we use the equation:

\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100

Putting values in above equation, we get:

\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%

The above calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

3 0
2 years ago
How many carbon atoms are there in a 1.3-carat diamond? diamonds are a form of pure carbon. (1 carat?
Colt1911 [192]
How many carbon atoms are there in a 1.3-carat diamond? Answer: 1.3 x 10^22 C atoms...
If a 1 carat = 0.20 g ... then 0.3 carat = 0.20 / 0.3 = 0.06 g
Thus, 1.3 carat = 0.26g
Find the moles first:
Moles= Grams / Mm of C
0.26 / 12.011 = 0.0216 mols of C
Atoms = Moles * Avogadro's number (6.022*10^23)
0.0216 * 6.022*10^23 = 1.3*10^22 C atoms

Hope this helps! :)
8 0
2 years ago
Read 2 more answers
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