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2 years ago

10

Draw the molecule on the canvas by choosing buttons from the tools(for bonds) atoms, and advanced template toolbars. The single

bond is active by default. Include all hydrogen atoms.I need help with solving this problem, thank you

1 answer:

Rasek [7]2 years ago

5 0

For this compound we will have this structure, describing the 3 - methyl - 1 - butyne

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lisov135 [29]

Answer:

Average atomic mass of uranium = 237.97 amu

Explanation:

Given data:

Abundance of U-234 = 0.01%

Abundance of U-235 = 0.71%

Abundance of U-238 = 99.28%

Average atomic mass of uranium = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

Average atomic mass = (234×0.01)+(235×0.71)+(238+99.28) /100

Average atomic mass = 2.34 + 166.85+23628.64 / 100

Average atomic mass = 23797.83 / 100

Average atomic mass = 237.97 amu.

3 0

3 years ago

Element X has an electron configuration that ends with 5p6. Which of the following describes this atom correctly?

pishuonlain [190]

The answer is d noble gas because the very last column on the periodic table is all the noble gases

7 0

4 years ago

If N gains 3 electrons from other atoms, why does it have a -3 charge?

muminat

Explanation:

Electrons are negatively charged. So it has a -3 charge.

3 0

3 years ago

The boiling point of bromine is 59 °C. Which of the following best predicts the boiling point of iodine monochloride, a polar co

soldier1979 [14.2K]

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

I just took the test and i got it right

6 0

4 years ago

Read 2 more answers

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 = $116^{0}C$

7 0

3 years ago