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4 years ago

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

Chemistry

2 answers:

tamaranim1 [39]4 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]4 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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Formation of magnesium chloride

Shkiper50 [21]

-Pure magnesium is commonly made by separating it from seawater. This process is known as electrolysis. The liquid magnesium formed is cooled into convenient blocks of metal known as ingots. The chlorine gas is recycled to form hydrochloric acid for the production of more magnesium chloride.

5 0

3 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago

What should be ph-value to be strong acid?

irakobra [83]

what should be ph-value to be strong acid?

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6 0

3 years ago

Read 2 more answers

The radius of a xenon atom is 1.3×10−8cm. a 100-ml flask is filled with xe at a pressure of 1.2 atm and a temperature of 281 k .

vladimir2022 [97]

Radius of Xenon = 1.3Ă—10â’8 cm
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288

3 0

4 years ago

Find the volume of a gas if 3.6 mols of the gas is at STP

Digiron [165]

Answer:

81 L gas

General Formulas and Concepts:

Ideal Gas Law

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.6 mols gas at STP

[Solve] volume (L) of gas

Step 2: Convert

[DA] Set up: $\displaystyle 3.6 \ mol \ gas(\frac{22.4 \ L \ gas}{1 \ mol \ gas \ at \ STP})$
[DA] Multiply [Cancel out units]: $\displaystyle 80.64 \ L \ gas$

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs.

80.64 L gas ≈ 81 L gas

6 0

3 years ago