3 years ago

What did ancient astronomers think areas of the moon called mares might be?

Chemistry

1 answer:

RUDIKE [14]3 years ago

7 0

Answer: seas

Explanation:

The ancient astronomers thought the mares on the moon were seas.

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The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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3 years ago

What can you say about the amount of electrons found in the second ring

Amiraneli [1.4K]

Answer:

The second ring in an atom can only hold up to 8 electrons.

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How many grams of Mg(ClO3)2 are needed to make 1,500 mL of a 2.5 N solution ?

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Explanation:

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2 years ago

Why is it that 85.48 rounded to two significant figures is 85 and not 86?

Gelneren [198K]

Answer:

See below.

Explanation:

That is because of the .48.

85.48 is closer to 85 than 86.

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3 years ago

a rock is found to have a small amount of copper would you call the rock an ore of copper explain in 5-6 sentences

alexgriva [62]

Answer:

Answer: No. An ore is a rock having large amounts of a mineral. Minerals may or may not be commercially useful.

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