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3 years ago

12

Helium only has two valence electrons and is a stable atom. How is this possible when atoms

1 answer:

Andru [333]3 years ago

6 0

Answer:

Explanation:

Helium (He) is similar in that it, too, only has room for two electrons in its only valence shell. Hydrogen and helium have only one electron shell. The first shell has only one s orbital and no p orbital, so it holds only two electrons. Therefore, these elements are most stable when they have two electrons.

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How many grams of O2(g) are needed to completely burn 46.0 g C3H8(g)?

Veronika [31]

Answer:

167,3 grams of O2 (g)

Explanation:

According to the following balanced equation:

C3H8 + 5 02 ---> 3 CO2 + 4 H20

We calculate the mass of the moles:

5 mol O2 = 16grams/mol x2x 5= 160 grams/mol

1 mol C3H8= 3 x12 grams/mol + 8 x 1 gram/mol= 44 grams/mol

If 44 grams C3H8 react with--------- 160 grams O2

46 gramsC3H8 react with --------X = (46x160)/44 =167, 3 grams O2

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4 years ago

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3 years ago

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3 years ago

Find the boiling point temperature at 760 torr of an isomer of octane, c8h18, if its enthalpy of vaporization is 38,210 j mol-1

mixas84 [53]

According to Clausius-Clayperon equation,

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

$P_{1}$ is the vapor pressure at boiling point = 760 torr

$P_{2}$ is the vapor pressure at T_{2} =638.43 torr

Temperature $T_{2} = 110.0^{0}C + 273 = 383 K$

Δ $H_{vap} = 38210 J/mol$

Plugging in the values, we get

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

ln $\frac{760 torr}{638.43 torr} =\frac{38210 J/mol}{8.314 J/(mol.K)} (\frac{1}{383 K}-\frac{1}{T_{1} }$

$T_{1} = 389 K$

Therefore, the boiling point of octane = 389 K - 273 = $116^{0}C$

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