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2 years ago

Which natural polymers are involved with making clothing?

Chemistry

2 answers:

fredd [130]2 years ago

7 0

Answer:

A. wool, silkworm, cocoon, and cellulose

Explanation:

I hope this helped!

Rina8888 [55]2 years ago

7 0

Answer:

I think from the choices that correct answer would be wool, silkworm cocoon, and cellulose. These are commonly used materials in clothing industry. Also, I don;t think that DNA would be used in making clothes. So, it should be the first option.

-Hope this helped

Explanation:

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What is the number of moles of 0.0960g of H2SO4

Vilka [71]

Answer=.000978802802moles H2SO4

How I Got My Answer

Molar mass
H2SO4= 98.079g/mol

What I have
.0960g H2SO4

Equation
.0960g*1mol/98.079g= .000978802802mol H2SO4

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2 years ago

Consider four sealed, rigid containers with the following volumes: 50 mL, 100 mL, 250 mL, and 500 mL. If each of these contains

dsp73

First, we assume that helium behaves as an ideal gas such that the ideal gas law is applicable.
PV = nRT
where P is pressure, V is volume, n is number of moles, R is universal gas constant, and T is temperature. From the equation, if n, R, and T are constant, there is an inverse relationship between P and V. From the given choices, the container with the greatest pressure would be the 50 mL.

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2 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

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2 years ago

With the help of balanced chemical equation explain what happens when:(1)Zinc is put in dilute hydrochloric acid (2)Zinc is put

UkoKoshka [18]

Find the attached document.