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3 years ago

Give three examples, from the lab, where potential energy was converted to kinetic energy.

Chemistry

1 answer:

GREYUIT [131]3 years ago

4 0

Give 3 Examples of where potential energy was converted to knlinetic energy:

Curtain

A ball before moving

An apple from the tree then falling down

When the Curtains are still, we call the that potential energy. If you move the curtains around, that is kinetic energy

The ball is still, that is potential energy. Then the ball is moving, the is kinetic energy

There is a apple ganging from a tree, that is potential energy. That apple is fall, this is kinetic energy

Hope this helps

Don't type or write in the answer, I'm not sure what from the lab means. These are a few potential into kinetic energy I could have think of!

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How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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3 years ago

What's the momentum of a 3.5-kg boulder rolling downhill at 5 m/s? A. 17.5 km·m/s B. 70 kg·m/s C. 8.5 kg·m/s D. 14.3 kg·m/s

swat32

Hey There!:

p = m * V

p = 3.5 * 5

p = 17.5 Km.m/s

Answer A

hope this helps!

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What process is required for minerals to crystallize from magma

OleMash [197]

Answer:

Chemical differentiation

Explanation:

The process by which minerals crystallize out of magma is termed chemical differentiation. Some times, it is referred to as magmatic differentiation.

Minerals have different properties and at certain environment, it comfortable for the settle out of magma.
Some minerals can crystallize out at very high temperature.
Some will form at medium temperature whereas, some are low temperature minerals.
Based on the properties of the minerals in a melt, a minerals will differentiate at various temperature and pressure regimes.

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Answer:

it helped alot of dcientisd

Explanation:

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A science teacher dissolved 50 grams of salt in 300 milliliters of water. he set the solution aside until all the water had evap

atroni [7]

When water evaporates, what is left behind is the salt that is dissolved. So you end up with the same amount of salt before and after. Meaning that the answer is 50 grams.

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3 years ago

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