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3 years ago

Heyyy, please help, I only need the sentences for the words please. Thank uuu :)

Chemistry

2 answers:

Trava [24]3 years ago

4 0

Answer:

Astronomical unit

a unit of measurement equal to 149.6 million kilometers, the mean distance from the center of the earth to the center of the sun.

Light year

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 trillion miles).

Planet

a celestial body moving in an elliptical orbit around a star.

Solar system

the collection of eight planets and their moons in orbit around the sun, together with smaller bodies in the form of asteroids, meteoroids, and comets. The planets of the solar system are (in order of distance from the sun) Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.

Galaxy

a system of millions or billions of stars, together with gas and dust, held together by gravitational attraction.

Universe

all existing matter and space considered as a whole; the cosmos. The universe is believed to be at least 10 billion light years in diameter and contains a vast number of galaxies; it has been expanding since its creation in the Big Bang about 13 billion years ago.

Star

a fixed luminous point in the night sky which is a large, remote incandescent body like the sun.

Explanation:

g100num [7]3 years ago

4 0

Answer:

Hello, riddle: What is the sound of a one hand clap.

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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Misha Larkins [42]

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D snails, guppies and plants. The community should be diverse and have lots of biodiversity.

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Travka [436]

Answer:

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Explanation:

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