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3 years ago

10

Helium has a density of 1.79×10^-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate th

e mass of helium that it would take to fill the balloon. Be sure to follow significant figure rules when calculating the answer.
A. 35,000g
B. 1.1×10^-3 g
C. 2.8×10^-5 g
D. 1.1g

1 answer:

djyliett [7]3 years ago

7 0

6,3 L = 6300 mL
..................
1,79×10^-4g -------- 1 mL
Xg --------------------- 6300 mL
X = 0,000179×6300
X = 1,1277g ≈ 1,1g

:•)

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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

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Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

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Given that:

the equilibrium number of vacancies at 800 °C

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Let's first determine the number of atoms in silver

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$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

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$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

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$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

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$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

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1 eV = 1.602176565 × 10⁻¹⁹ J

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$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

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According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

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$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

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Moles of hydrogen gas obtained experimentally = 1.1607 mol

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