All categories

3 years ago

A scientist discovers a new type of polymer. What must be true about this polymer?

Chemistry

1 answer:

DochEvi [55]3 years ago

3 0

The correct answer is D

a polymer is a substance made up of large molecules with repeating sub units called monomers.

A is incorrect because its a ring structure which means it cannot form the long chain that is exhibited by polymers.

B is incorrect because the double bonds will cause saturation, hence no monomer can bond to another monomer to form the large molecule.

C is incorrect because the large molecule has repeating sub-units hence only specific types of molecules can be present, and it is usually carbon and hydrogen.

You might be interested in

Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.

lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i) 0.115M 0 0

ΔC -x +x +x

C(eq) 0.115M - x x x

≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0

3 years ago

When you put an ice cube into a glass of room temperature water, why does the water get cold?

mel-nik [20]

The water gets cold because the ice cube is endothermic. Which means it takes in heat. It takes the warmth from the water and melts the ice cube which the ice cube releases coldness which cools down the water.

6 0

3 years ago

The head loss in a turbulent flow in a pipe varies Approximant as square of velocity • Direct as the velocity • Invers as square

Eddi Din [679]

Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

$h_F=\dfrac{FLV^2}{2gD}$

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

$h_F\alpha V^2$

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

5 0

3 years ago

Read 2 more answers

Please help me with this question!!

mestny [16]

Answer: 824.6 g of NaCl are produced from 500.0 g of chlorine.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$ $\text{Moles of} Cl_2=\frac{500.00g}{71g/mol}=7.04moles$

$2Na+Cl_2\rightarrow 2NaCl$

According to stoichiometry :

1 mole of $Cl_2$ produce = 2 moles of $NaCl$

Thus 7.04 moles of $Cl_2$ will produce= $\frac{2}{1}\times 7.04=14.08moles$ of $NaCl$

Mass of $NaCl=moles\times {\text {Molar mass}}=14.08moles\times 58.5g/mol=824.26g$

Thus 824.6 g of NaCl are produced from 500.0 g of chlorine.

3 0

3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago