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3 years ago

How to stimulate softness?

Chemistry

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

factors affect how human touch perceives softness, like the feel of pressing your fingertip against a marshmallow, a piece of clay or a rubber ball? By exploring this question in detail, a team of engineers and psychologists at the University of California San Diego discovered clever tricks to design materials that replicate different levels of perceived softness.

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The walls in your home have 25mm of expanded polystyrene (R-value 3.96) and 25mm of rigid urethane foam (R-value 7.50). What is

Finger [1]

Answer:

3.6 per inch

EPS insulation features an average r-value of 3.6 per inch.

In addition to providing energy efficiency, EPS without film facers has a perm rating of up to 5.0.

Explanation:

hope this helps

7 0

4 years ago

Which scenario involves a reaction that is at equilibrium?

musickatia [10]

Answer:

the amount of products and reactants is constant

4 0

3 years ago

प्रश्न 5. 'पुष्प की अभिलाषा' कविता में पुष्प के दवारा क्या अभिलाषा व्यक्त की गई

Kay [80]

Please ask in English

4 0

3 years ago

ammonia (NH3(g) Hf=-45.9 kJ/mol) reacts. with oxygen to produce nitrogen and water (H2O(g) Hf = -241.8 kJ/mol according to the e

kherson [118]

Answer:

ΔH°_rxn = -195.9 kJ·mol⁻¹

Explanation:

4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)

ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8

The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ

ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ

6 0

3 years ago

Read 2 more answers

The vapor pressure of pure water at 296 K is 2778.5 Pa. The vapor forms an ideal gas. 1) In some oil, the equilibrium concentrat

Murljashka [212]

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, $\text{(Vapor pressure)}_{mixture}$ = $\frac{1}{100} \times 2778.5 Pa$

= 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.

3 0

3 years ago