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2 years ago

Real life example of when temperature increases, then the pressure increases.

Chemistry

1 answer:

timurjin [86]2 years ago

6 0

Answer:

Tyre

In cold weather, you might have regularly kept a check on the pressure of the tyres of your car. Driving increases the temperature of the tyres, and, therefore, the air inside the tyre warms and expands. When you measure the pressure of the tyres at the time when you have just driven the car, it will be high. However, in cold weather, the pressure of the tyres will be low. So, it is recommended that you should always measure the pressure of the tyres.

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What is the function of a Mitochondrion

gladu [14]

Answer:

Explanation:

Function. The mitochondrion is the site of ATP synthesis for the cell. The number of mitochondria found in a cell are therefore a good indicator of the cell's rate of metabolic activity; cells which are very metabolically active, such as hepatocytes, will have many mitochondria.

6 0

2 years ago

What kind of solution would have a Kb value much greater than 1?

Marta_Voda [28]

Answer:

B. a strongly basic solution

Explanation:

Kb is base dissociation constant, which indicates how completely a base dissociates into its component ions in water. The greater the Kb value, the greater the alkalinity of the solution and vice versa.

Therefore, a solution with a Kb value much greater than 1, indicates a strongly basic solution, while a solution with a Kb value less than 1, indicates a weakly basic solution.

8 0

2 years ago

Antarctica, almost completely covered in ice, has an area

Marysya12 [62]

Answer: The mass of ice is $2.39\times 10^{22}g$

Explanation:

We are given:

Area of Antarctica = $5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2$ (Conversion factor: $1mi^2=2.59\times 10^{10}cm^2$ )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = $182.88\times 10^3cm$ (Conversion factor: 1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of ice = $0.917g/cm^3$

Volume of ice = Area × Height of ice = $142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3$

Putting values in above equation, we get:

$0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g$

Hence, the mass of ice is $2.39\times 10^{22}g$

5 0

2 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

laila [671]

Answer:

First choice: 99 J/K

Explanation:

1) First law of thermodynamic (energy balance)

Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

2) Energy change of each substance:

General formula:

Heat released or absorbed = mass × Specific heat × change in temperature

density of water: you may take 0.997 g/ ml as an average density for the water.

mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

Specif heat of water: 1 cal / g°C

Heat released by the hot water:

Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

Heat absorbed by the cold water:

Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

Heat absorbed by the calorimeter

Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

4) Balance

Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

Convert to cal / K to Joule / K

1 cal = 4.18 Joule

23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0

3 years ago

Write one other example (other than the one in your note) of ore of the following metals

borishaifa [10]

An exemple of aluminium in its ore state is haemetite

6 0

2 years ago