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3 years ago

What volume of lead (of density 11.3 g/cm3) has the same mass as 330 cm3 of a piece of redwood (of density 0.38 g/cm3)?

Chemistry

1 answer:

zimovet [89]3 years ago

3 0

Answer:

The sample of lead has a volume of 11.1 cm³

Explanation:

Step 1: Data given

x cm³ lead has a density of 11.3 g/cm³

it has the same mass as 330cm³ of a piece of redwood with density 0.38g/cm³

Step 2: Calculate mass of the piece of redwood

Density = mass/volume

mass = density * volume

Mass of the piece of redwood = 0.38 g/cm³ * 330cm³ = 125.4 grams

Since the sample of lead has the same mass, it also has a mass of 125.4 grams

Step 3: Calculate volume of the lead

Density = mass/ volume

Volume = mass/ density

Volume of lead = 125.4g / 11.3g/cm³ = 11.097 cm³≈11.1 cm³

The sample of lead has a volume of 11.1 cm³

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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]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

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

For b:

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

C3H8 + 5 O2 → 3 CO2 + 4 H2O What is the number of atoms on each side of the equation?

Roman55 [17]

Answer: 1.2642*10²⁵ on both sides

Explanation:

First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides

There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21

Now use Avogrado's constant

21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³

= 1.2642*10²⁵

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