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Aleks [24]
2 years ago
8

What is the pressure in a 27.0-L cylinder filled with 44.9 g of oxygen gas at a temperature of 315 KK

Chemistry
1 answer:
steposvetlana [31]2 years ago
6 0

Answer:

1.343 atm

Explanation:

We are given the following;

Pressure, p = ?

Volume v = 27 L

Mass of oxygen = 44.9 g

Temperature, T = 315 K

The formular relating all these variables is the equation;

PV = nRT

where R = gas constant = 0.08206 L atm / mole K

To obtain n, we use;

number of moles, n = Mass /  molar mass =  44.9 / 32 = 1.403 moles

From the ideal gas equation;

P = nRT / V

P = 1.403 * 0.08206 * 315 / 27

P = 36.27 / 27 = 1.343 atm

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Which of the following substances is formed during photosynthesis A.carbon dioxide
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A block of aluminum occupies a volume of 15.0 cm and weighs 40.5 g. What is its density?
mafiozo [28]

Answer:

2.7 g/cm^3

Explanation:

D = m/V

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An ion of oxygen- 16 contains 8 protons and has a 2- charge. How many electrons does it have?
Verdich [7]

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i would say 10, so the anser is A.

Explanation:

because there are the same number of protons and electrons, therefore for a regular O, you are supposed to have only 8 protons, but it is charged, thus, whatever the charge is will be taken into consideration into how much the proton and electron doe it have. Thus, for this case, it has 10, because the charge is negative and you have 8 electron plus 2 = 10.

3 0
2 years ago
Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6
MAXImum [283]

Answer:

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

Explanation:

Given that:

the equilibrium  number of vacancies at 800 °C

i.e T = 800°C     is  3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

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

where ;

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

\rho _{Ag} = Density of silver = 9.5 g/cm³

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

However;

The equation for equilibrium number of vacancies can be represented by the equation:

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}

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

Q_v =  (2.38 \ * 1.602176565 * 10^{-19} ) J/atom  }

\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}

8 0
3 years ago
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