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Sindrei [870]
2 years ago
8

What tool would you use to find the mass of 10 marbles?​

Chemistry
1 answer:
erastovalidia [21]2 years ago
7 0

Answer:

I think you would use a balance!

Explanation:

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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

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

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
HELP
tatiyna

Answer:

Bi (Bismuth)

Ag (Silver)

Li (Lithium)

Explanation:

Xe (Xenon) and I (Iodine) are non-metals. They cannot from a metallic bond because metallic bonds are bonds between metals only.

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2 years ago
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natita [175]

Answer:

Explanation:

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5 0
3 years ago
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Calculate the energy(J) change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom. Put answer in sci
Oliga [24]
The energy at n level of hydrogen atom energy level =13.6/n^2

substiture the respective n values in the equation above and find the difference in the energy levels

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3 0
3 years ago
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A gas has a pressure of 1.34 atm when the temperature is 237K. The gas is then heated until the temperature measures 312K. What
alexandr1967 [171]

The answer for the following question is answered below.

  • <em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

Explanation:

Given:

Initial pressure of the gas = 1.34 atm

Initial temperature of the gas = 273 K

final temperature of the gas = 312 K

To solve:

Final temperature of the gas

We know;

From the ideal gas equation

P × V = n × R × T

So;

from the above equation we can say that

    <em>P ∝ T</em>

     \frac{P}{T} = constant  

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

Where;

P_{1} = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of  a gas

    P_{2} = \frac{1.34*312}{237}

    P_{2}  = 1.76 atm

<em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

6 0
3 years ago
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