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Amiraneli [1.4K]
3 years ago
9

HELP PLEASE

Chemistry
1 answer:
gavmur [86]3 years ago
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.

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What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?
Zigmanuir [339]

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

Learn more about:

molar concentration

brainly.com/question/1286583

#learnwithBrainly

6 0
3 years ago
2. Which of the following expressions is generally used for solubility? (1 point)
miskamm [114]

<u>Answer:</u>

For 2: The correct answer is grams of solute per 100 grams of solvent.

For 3: The correct answer is supersaturated.

For 4: the correct answer is the solubility decreases.

<u>Explanation:</u>

  • <u>For 2:</u>

Solubility is defined as the property which refers to the ability of the solute that can be dissolved in a solvent. It is defined as the number of grams of  solute per 100 grams of solvent.

  • <u>For 3:</u>

Unsaturated solution is defined as the solution in which amount of solute that is dissolved in the solvent is less.

Saturated solution is defined as the solution in which no more solute can be dissolved in the given amount of solvent.

Emulsion is defined as the dispersion of one liquid in another liquid in which it is not soluble.

Supersaturated solution is defined as the solution in which solvent contains more amount of solute than the required amount. These solutions help in the process of crystallization.

When a crystal is added to a <u>supersaturated solution</u>, more and more particles come out of the solution and this process is known as crystallization.

  • <u>For 4:</u>

According to the Henry's Law

The solubility of the gas in a liquid is directly proportional to the partial pressure of the gas.

\text{Solubility of a gas in a liquid}\propto \text{Partial pressure of the gas}

With increase in the partial pressure, the solubility of the gas in liquid also increases and vice-versa.

Hence, the correct answer is the solubility decreases.

7 0
3 years ago
Read 2 more answers
A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T
taurus [48]

Answer:

<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

Explanation:

  • Adding solute to water causes the depression of the freezing point.

  • We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

  • <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

6 0
3 years ago
The pattern of colors given off by a particular atom is called
babymother [125]

Answer:

idk

Explanation:

3 0
3 years ago
If you have 85.0 grams of CO₂, calculate how many moles of CO₂ would you have?
dedylja [7]
N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles
4 0
2 years ago
Read 2 more answers
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