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2 years ago

How do we measure the mass of an extrasolar planet?

1 answer:

4 0

To determine a planet's mass, astronomers typically measure the minuscule movement of the star caused by the gravitational tug of an orbiting planet. For planets the massof Earth detecting such a tiny tug is extraordinarily challenging with current technology

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Which of the following is not observed in a homologous series?

Stolb23 [73]

Answer:

Change in chemical properties

Explanation:

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2 years ago

Which statement describes a reason to consider air a mixture? A) The major constituents of air are gaseous elements. B) The comp

umka21 [38]

The correct answer is - A) The major constituents of air are gaseous elements.

With the statement ''the major constituents of air are gaseous elements'' we can easily conclude that the air is a mixture. The reason for that is that we have a plural usage of the word element, elements, which mean that there are multiple elements that make up the air.

The air is indeed predominantly a mixture of gaseous elements. The most abundant gas in the air being the nitrogen with 78.9%, oxygen with 20.95%, argon 0.93%, and carbon dioxide 0.04%, with lesser amounts of other gases also be present in it. The water vapor is also present in the air, though it is variable, being around 1% at sea level, but only 0.4% over the entire atmosphere.

3 0

3 years ago

4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$