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2 years ago

9

Exactly 1 mol Hg2(NO3)2 contains how many moles of Hg , N , and O ?

1 answer:

marin [14]2 years ago

6 0

In exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Since the molecular formula of Hg₂(NO₃)₂ shows that for every mole of the substance, we have 2 moles of Hg, 2 moles of N and 6 moles of O.

So, in exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Learn more about number of moles here:

brainly.com/question/3935424

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3 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

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Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

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3 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

yan [13]

Answer:

Water will boil at $76^{0}\textrm{C}$ .

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

$ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$

where, $P_{1}$ and $P_{2}$ are vapor pressures of liquid at $T_{1}$ (in kelvin) and $T_{2}$ (in kelvin) temperatures respectively.

Here, $P_{1}$ = 760.0 mm Hg, $T_{1}$ = 373 K, $P_{2}$ = 314.0 mm Hg

Plug-in all the given values in the above equation:

$ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]$

or, $T_{2}=349 K$

So, $T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}$

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3 years ago