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Karo-lina-s [1.5K]
3 years ago
7

Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for p

acking. For example, polonium has a density of 9.19 g/cm3 and a unit cell side length a of 3.37 Å. (1 Å = 1 10-8 cm.) (a) How many polonium atoms are in exactly 1 cm3? atoms (b) How many unit cells are in exactly 1 cm3? unit cells (c) How many polonium atoms are there per unit cell?
Chemistry
1 answer:
jarptica [38.1K]3 years ago
3 0

Answer:

A.) 34.866 x 10¯23 g/atom

B.) 2.64 x 10^22 atoms in 1 cm3

C.) 1 atoms per unit cell

Explanation:

A) Calculate the average mass of one atom of polonium:

210g mol¯1 ÷ 6.022 x 1023 atoms mol¯1 = 34.866 x 10¯23 g/atom

B) Determine atoms in 1 cm3:

9.19g / 34.866x 10¯23 g/atom = 2.64 x 10^22 atoms in 1 cm3

Determine volume of the unit cell:

(3.37 x 10¯8 cm)3 = 3.827 x 10¯23 cm3

Determine number of unit cells in 1 cm3:

1 cm3 / 3.827 x 10¯23 cm3 = 2.61 x 1022 unit cells

C) Determine atoms per unit cell:

2.64 x 1022 atoms / 2.61 x 1022 unit cells = 1 atoms per unit cell

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A gas that was cooled to 200 Kelvin has a volume of 65.8 L. If its initial volume was 132.4 L, what was its initial temperature?
Mandarinka [93]

Answer:

Initial temperature, T1 = 99.4 Kelvin

Explanation:

<u>Given the following data;</u>

  • Initial volume, V1 = 65.8 Litres
  • Final temperature, T2 = 200 Kelvin
  • Final volume, V2 = 132.4 Litres

To find the initial temperature (T1), we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

\frac {V}{T} = K

\frac {V_{1}}{T_{1}} = \frac {V_{2}}{T_{2}}

Making T1 as the subject formula, we have;

T_{1} = \frac {V_{1}T_{2}}{V_{2}}

Substituting the values into the formula, we have;

T_{1} = \frac {65.8 * 200}{132.4}

T_{1} = \frac {13160}{132.4}

<em>Initial temperature, T1 = 99.4 Kelvin</em>

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3 years ago
A 34.4 L sample of oxygen gas at 229°C and 752 torr is cooled to 34°C at 668 torr. The volume of the sample is now
kenny6666 [7]

Answer:

23.55 L

Explanation:

USe the following 'identity' of gs laws

P1 V 1 / T1 = P2 V2 / T2         ( T must be in Kelvin)

re arrange to

P1 V 1  T2  /  (T1 P2)   = V2       NOW SUB IN THE VALUES

752 * 34.2 * ( 34 + 273.15) / [( 229 + 273.15) * 668]   = V2 = 23.55 L

5 0
1 year ago
At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N
zhenek [66]

Answer : The value of equilibrium constant (K) is, 5.71\times 10^4

Explanation :

First we have to calculate the concentration of N_2,O_2\text{ and }N_2O

\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M

and,

\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

N_2(g)+O_2(g)\rightarrow 2N_2O(g)

The expression for equilibrium constant is:

K=\frac{[N_2O]^2}{[N_2][O_2]}

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

K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}

K=5.71\times 10^4

Thus, the value of equilibrium constant (K) is, 5.71\times 10^4

7 0
3 years ago
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