Why is it NOT possible for ONE particle (atom/ion/molecule) to be classified as a particular phase, or state, of matter?

3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.

tiny-mole [99]

Answer:

The Avogadro's number is $N_A = 6.02289 *10^{23}$

Explanation:

From the question we are told that

The edge length is $L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}$

The density of the metal is $\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3$

The molar mass of Ba is $Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol$

Generally the volume of a unit cell is

$V = L^3$

substituting value

$V = [5.02 *10^{-10}]^3$

$V = 1.265*10^{-28}\ m^3$

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),

The volume of barium atom is

$V_a = \frac{V}{2} * 0.68$

substituting value

$V_a = \frac{ 1.265*10^{-28}}{2} * 0.68$

$V_a = 4.301 *10^{-29} \ m^3$

The Molar mass of barium is mathematically represented as

$Z = N_A V_a * \rho$

Where $N_A$ is the Avogadro's number

So

$N_A = \frac{ Z}{ V_a * \rho}$

substituting value

$N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}$

$N_A = 6.02289 *10^{23}$

4 0

2 years ago

Use the ideal-gas law to estimate the number of air molecules in your physics lab room, assuming all the air is N2. Assume a roo

stiv31 [10]

Answer:

4.13×10²⁷ molecules of N₂ are in the room

Explanation:

ideal gases Law → P . V = n . R . T

Pressure . volume = moles . Ideal Gases Constant . T° K

T°K = T°C + 273 → 20°C + 273 = 293K

Let's determine the volume of the room:

18 ft . 18 ft . 18ft = 5832 ft³

We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L

1 atm . 165045.6 L = n . 0.082 L.atm/mol.K . 293K

(1 atm . 165045.6 L) / 0.082 L.atm/mol.K . 293K = n

6869.4 moles of N₂ are in the room

If we want to find out the number of molecules we multiply the moles by NA

6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

4 0

3 years ago

The rate constant for the decomposition of nitrogen dioxide NO2(g) LaTeX: \longrightarrow⟶ NO (g) + 1/2 O2(g) with a laser beam

aleksley [76]

Answer :

The time taken by the reaction is 19.2 seconds.

The order of reaction is, second order reaction.

Explanation :

The general formula to determine the unit of rate constant is:

$(Concentration)^{(1-n)}(Time)^{-1}$

Unit of rate constant Order of reaction

$Concentration/Time$ 0

$Time^{-1}$ 1

$(Concentration)^{-1}Time^{-1}$ 2

As the unit of rate constant is $M^{-1}min^{-1}$ . So, the order of reaction is second order.

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $1.95M^{-1}s^{-1}$

t = time = ?

$[A_t]$ = final concentration = 0.97 M

$[A_o]$ = initial concentration = 2.48 M

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

$1.95\times t=\frac{1}{0.97}-\frac{1}{2.48}$

$t=0.32min=0.32\times 60s=19.2s$

Therefore, the time taken by the reaction is 19.2 seconds.

8 0

2 years ago

What contribution to atomic theory resulted from albert einstein’s work?

gizmo_the_mogwai [7]

Answer: A new model of the atom that described electrons as being in a cloud

Explanation:

7 0

3 years ago

Read 2 more answers

Someone help me pleasee?

Kamila [148]

Maybe its the second one

7 0

3 years ago