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4 years ago

HELP PLEASE

1 answer:

8 0

There are a number of ways to reduce friction:

Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction. Some swimmers wear suits to reduce underwater resistance. These suits mimic the smooth skin of sharks.

Lubrication is another way to make a surface smoother. A lubricant is a slippery substance designed to reduce the friction between surfaces. You might use oil to stop a door from squeaking - the oil reduces the friction in the hinge. Water can be used as a lubricant - think of how a floor becomes slippery after it has been mopped.

Make the object more streamlined. A streamline shape is one that allows air or water to flow around it easily, offering the least resistance. Compare a boxy old car with a new car that has a rounded shape, allowing it to move with less effort.

Reduce the forces acting on the surfaces. The stronger the forces acting on the surfaces, the higher the friction, so reducing the forces would reduce the friction. If you apply the handbrake when you try to drive a car, the car will have a lot of difficulty moving because of the force immobilising (stopping the movement of) the wheels. If you release the handbrake, the wheels will move more freely because there is no extra force acting on them.

Reduce the contact between the surfaces. Have you ever tried to roll a cube? Spheres are the best shape for reducing friction because very little of a spherical object is in contact with the other surface. Several types of wheels, such as skateboard wheels, contain small spheres called ball bearings to reduce the friction between the moving parts. You can witness the effect of ball bearings by comparing the friction between sliding a book on a table and then doing the same, but using marbles between the book and the surface of the table. Notice how the marbles act as ball bearings, reducing the friction.

You might be interested in

What is the solubility in moles/liter for calcium phosphate at 25°C given a Ksp value of 1.2 x 10^-29. Write using scientific no

Oksanka [162]

Answer:

Explanation:

Ca₃( PO₄)₂ = 3 Ca⁺² + 2 PO₄⁻³

x 3x 2 x

Let solubility be x mole per litre

solubility product = (3x )³ x ( 2x )²

= 108 x⁵ = 1.2 x 10⁻²⁹

x⁵ = .0111 x 10⁻²⁹ = .111 x 10⁻³⁰

x = .64 x 10⁻⁶ moles / litre

6 0

3 years ago

Question 1 (Matching Worth 3 points)

Misha Larkins [42]

Answer:

Question 1: 1) Increasing the pressure C) Shift to the right

2) Removing hydrogen gas A) Shift to the left

3) Adding a catalyst B) No effect

Question 2: This reaction is exothermic because the system shifted to the right on cooling.

Question 3: Shift it toward the reactants.

Question 4: Adding more of gas C to the system.

Question 5: It will shift toward the reactant side because the reactant side has one more mole of gas than the product side.

Question 6: True.

Question 7: there is no suitable choice is provided.

We can shift the equilibrium toward the right via:

Increasing N2O3 concentration,

decreasing NO and/or NO2 concentration,

decreasing the pressure,

lowering the T (cooling the system).

Explanation:

Question 1: Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g).

Match Term Definition

Removing ammonia A) No effect

Removing hydrogen gas B) Shift to the left

Adding a catalyst C) Shift to the right

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

1) Increasing the pressure:

When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.

So, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).

so, the right match is: C) Shift to the right.

2) Removing hydrogen gas:

Removing hydrogen gas will decrease the concentration of the products side, so the reaction will be shifted to the lift side to suppress the decrease in the concentration of hydrogen gas by removing.

so, the right match is: A) Shift to the left.

3) Adding a catalyst:

Catalyst increases the rate of the reaction without affecting the equilibrium position.

Catalyst increases the rate via lowering the activation energy of the reaction.

This can occur via passing the reaction in alternative pathway (changing the mechanism).

The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).

in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.

with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.

so, the right match is: B) No effect.

Question 2: Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

2NO2(g) ⇌ N2O4(g)

When the light brown color equilibrium mixture was moved from room temperature to a lower temperature, the mixture turned lighter brown in color. Which of the following conclusions about this equilibrium mixture is true?

Moving from room temperature to a lower temperature, means that the T is decreased.

The mixture turned lighter brown in color, means that the reaction is shifted towards the products side.

Moving towards the product side, means that we decreased the concentration of the product side, which ,means that the heat is evolved from the reaction and the reaction is exothermic.

So, the right choice is: This reaction is exothermic because the system shifted to the right on cooling.

Question 3: According to Le Châtelier's principle, how will a decrease in concentration of a reactant affect the equilibrium system?

A decrease in the reactants concentration will disturb the equilibrium position of the system. To attain the equilibrium again, the system will be shifted to the lift (reactants side) to increases the concentration of the reactants and attain the equilibrium again.

So, the right choice is: Shift it toward the reactants.

Note: The answer of Q 4, 5, 6 & 7 and all answers are in the attached word file.

Download docx

5 0

4 years ago

How is stoichiometry used to calculate energy released when a mass of liquid freezes?

kati45 [8]

Answer:

D. Grams liquid x mol/g x delta Hfreezing

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to reason that the stoichiometry used to calculate energy released when a mass of liquid freezes, involves the grams of the liquid, the molar mass of the liquid, as given in all the group choices, and the enthalpy of freezing because that is the process whereby a liquid goes solid.

In such a way, we infer that the correct factor would be D. Grams liquid x mol/g x delta Hfreezing which sometimes is the negative of the enthalpy of fusion as they are contrary processes.

Regards!

6 0

3 years ago

Read 2 more answers

Chlorine melts at 172 K. What is chlorine’s melting point in °C?

Alexxx [7]

-101.5 celcius and exactly that

7 0

4 years ago

Read 2 more answers

The oxidation of an unsaturated fatty acid leads to the production of fewer ATPs than a saturated fatty acid with the same numbe

kirill [66]

Answer:

C) the double bond results in the bypassing of the first oxidation step of the pathway, thereby eliminating one FADH2 product

Explanation:

The beta oxidation of fatty acids is the process by which long chain saturated and unsaturated fatty acids are oxidized to the two-carbon product, acetylCoA.

The beta oxudation of fatty acids occurs in four steps after which the chain length is shortened by 2 with cleavage of an acetylCoA molecule.

The first step involves the oxidation of the fatty acid with introduction of a double bond between C-2 and C-3 of the long carbon chain. This reaction produces FADH2 which donates its electrons to electron carriers in the electron transport chain tomproduce 1.5 moles of ATP. This is the case with saturated fatty acids.

However, in unsaturated fatty acids (which are less reduced than saturated fatty acids) where there is already a double bond innthe molecule, the double bond results in the bypassing of the first oxidation step of the pathway, thereby eliminating one FADH2 product.

This causes a reduction In the number of ATP produced by unsaturated fatty acids when compared to saturated fatty acids.

8 0

3 years ago