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kicyunya [14]
3 years ago
7

Calculate density of aluminum if 27.6cm^3 has a mass of 74.6g

Chemistry
2 answers:
rjkz [21]3 years ago
6 0
First , 27.6 cm^3 is equal to 27.6 ml

Density = mass g / volume cm^3

= 74.6 / 27.6 = ........ g/cm^3
PolarNik [594]3 years ago
5 0

Answer : The density of aluminium will be, 2.73g/cm^3

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

Density=\frac{Mass}{Volume}

Given :

Mass of aluminum = 74.6 grams

Volume = 27.3cm^3

Now put all the given values in the above formula, we get the density of aluminium.

Density=\frac{74.6g}{27.3cm^3}=2.73g/cm^3

Therefore, the density of aluminium will be, 2.73g/cm^3

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A 3.00 L flexible container holds a sample of hydrogen gas at 153 kPa. If the pressure increases to 203 kPa and the temperature
dybincka [34]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

V2 = P1 V1 / P2

V2 = 153 x 3.00 / 203

<span>V2 = 2.26 L</span>

3 0
3 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
What is a mol in chemistry
Snowcat [4.5K]
Molecular weight it stands for molecular weight






6 0
4 years ago
What happens to the matter in the body of an animal after it dies
sladkih [1.3K]
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
4 0
3 years ago
What substances is a liquid at room temperature
Vesnalui [34]

Answer:

Only two elements are liquid at standard conditions for temperature and pressure: mercury and bromine. Four more elements have melting points slightly above room temperature: francium, caesium, gallium and rubidium.

Explanation:

6 0
3 years ago
Read 2 more answers
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