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denpristay [2]
3 years ago
6

A metal crystallizes in a body-centered cubic unit cell. The radius of one atom = 2.30 x 10 -8 cm. The density of the metal is 0

.867 g/cm^3 . What is molar mass of metal?
Chemistry
1 answer:
maw [93]3 years ago
6 0

Answer:

25.41 g/mol is molar mass of metal.

Explanation:

Number of atom in BCC unit cell = Z = 2

Density of metal = 0.867 g/cm^3

Edge length of cubic unit cell= a = ?

Radius  of the atom of metal = r = 2.30\times 10^{-8} cm

a=2\times r=2\times 2.30\times 10^{-8} cm=4.60\times 10^{-8} cm

Atomic mass of metal =M

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

0.867 g/cm^3=\frac{2\times M}{6.022\times 10^{23} mol^{-1}\times (4.60\times 10^{-8} cm)^{3}}

M = 25.41 g/mol

25.41 g/mol is molar mass of metal.

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A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the
cupoosta [38]

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules =T_1

Molecules helium gas = x

Moles of helium has = n_1= \frac{x}{N_A}

PV = nRT (Ideal gas equation)

PV=n_1RT_1...[1]  

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = \frac{x}{4}

Moles of helium has left after removal = n_2= \frac{x}{4\times N_A}

PV=n_2RT_2...[2]

n_1RT_1=n_2RT_2

\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2

T_1=\frac{T_2}{4}

T_2=4T_1

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

6 0
3 years ago
A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of th
nekit [7.7K]
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


3 0
3 years ago
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Poem about scientific method​
VashaNatasha [74]

Answer:

Here is the link

Explanation:

A poem about the scientific method

6 0
3 years ago
Earth has approximately 600,000,000 meters of coastline. If we assume this entire length of coastline has sandy beaches 60 meter
VashaNatasha [74]

Answer:

7 and 11

Explanation:

The amount of sand on the beaches can be found using this formula:

volume (m3) = length (m) × width (m) × depth (m)

(6 × 108 m) × 60 m × 20 m = 7 × 1011 m3

Therefore, there would be a total of 7 × 1011 cubic meters of sand on the beaches.

4 0
3 years ago
The North Star is the last star in the Ursa Minor constellation.<br> True<br> False
Alja [10]
The answer is true. It is the last star.
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