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andriy [413]
2 years ago
14

I will give brainliest. 1. A spiderweb and a Kevlar jacket have some obvious differences. Which property is similar between the

web and the jacket?
- surface area
- thickness, expressed as a number of atoms
- overall strength
- arrangement of atoms in molecules
2. Assuming silk from spiderwebs could be made just as strong as Kevlar, why would a company still choose to use Kevlar in producing bulletproof fabrics?
- A much larger amount of silk might be needed to produce the same effect.
- Spiderweb silk would likely be rejected by the body.
- The cost might be higher for producing spiderweb silk.
- Spiderweb silk likely involves more chemicals.
3. Which microscopic detail affects the strength of different forms of silk?
- Different silk strands are made with different types of atoms.
- Different types of silk are in long strands or in other very different arrangements.
- Different silk strands have different combinations of amino acids.
- Different types of silk do not have the same strength if they come from different sources.
4. Which type silk is the strongest?
- nonbiodegradable
- minor ampullate
- major ampullate
- biodegradable
5. Which natural source has been bioengineered to make silk proteins?
- mulberry leaves
- goat milk
- goat hair
- mulberry bark
Chemistry
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

  • arrangement of atoms in molecules

  • .A much larger amount of silk might be needed to produce the same effect.

  • - Different silk strands have different combinations of amino acids.

  • .major ampullate
  • .goat milk

Explanation:

Using Organic Compounds quick check

do you know the refining design quick check

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A discarded spray paint can contains only a small volume of the propellant gas at a
Katena32 [7]

Answer:

1.24 × 10³ kPa

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 34.5 kPa
  • Initial volume of the can (V₁): 473 mL
  • Final pressure of the gas (P₂): ?
  • Final volume of the can (V₂): 13.16 mL

Step 2: Calculate the final pressure of the gas in the can

If we assume that the gas in the can behaves as an ideal gas and that the temperature remains constant, we can calculate the final pressure of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 34.5 kPa × 473 mL / 13.16 mL = 1.24 × 10³ kPa

5 0
3 years ago
I need to know 5 and 6
irina [24]
6-cumbustion reaction
7 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
4 years ago
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