D. What do all of the molecules in the table have in common?

Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c

inysia [295]

First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams

Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

Zinc mass = 0.37 x 954
Zinc mass = 352.98 g
Zinc moles = 352.98 / 65
Zinc moles = 5.43
Zinc atoms = 5.43 x 6.02 x 10²³
Zinc atoms = 3.27 x 10²⁴

Copper mass = 0.63 x 954
Copper mass = 601.02 g
Copper moles = 601.02 / 64
Copper moles = 9.39
Copper atoms = 9.39 x 6.02 x 10²³
Copper atoms = 5.56 x 10²⁴

3 0

2 years ago

Consider the following operations on the number 7.02 times 10^-2. Without using a calculator, decide which would give a signific

mariarad [96]

Answer: (a). Significantly Larger value

(b). Significantly Smaller value

(c). Significantly Larger value

(d). Significantly Smaller value

Explanation:

This is quite a dicey question to analyze, let us not get too carried away with the simplicity of the mathematical signs involved.

Given that 7.02 × 10⁻² is the number to be compared with.

(a). 7.02 × 10⁻² + 6.10 × 10⁵

from this, we can see a very large number being added to the base number involved, although the base number is carrying a negative power, it does not affect the other greatly.

ANS: the sum would give a significantly larger value

(b). 7.02 × 10⁻² - 6.10 × 10⁵

Here, we can see a very large number subtracted from the base number having a negative power.

the subtraction of this numbers gives a Significantly smaller value than the base number i.e. 7.02 × 10⁻²

ANS: Significantly smaller value

(c). 7.02 × 10⁻² × 6.10 × 10⁵

we would solve this from our basic knowledge of indices

where ⇒ (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ)

we have,

7.02 × 10⁻² × 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁺⁵ = (7.02 × 6.10) × 10³

this final value (multiplication value) gives a Significantly larger value.

ANS: Significantly larger value

(d). 7.02 × 10⁻² ÷ 6.10 × 10⁵

Also, we apply the indices rule for Division

where ⇒ (10ᵃ ÷ 10ᵇ = 10ᵃ⁻ᵇ)

i.e. 7.02 × 10⁻² ÷ 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁻⁺⁵ = (7.02 × 6.10) × 10⁻⁷

this final value gives a Significantly Smaller value

ANS: Significantly Smaller value

cheers i hope this helps.

7 0

3 years ago

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

Help me answer this question pls

Nostrana [21]

The answer should be D all of the above

3 0

3 years ago

Read 2 more answers

Vanadium (II) oxide with Iron (III) oxide results in Vanadium (V) oxide and Iron (II) oxide. Balance this equation

evablogger [386]

Answer:

2VO + 3Fe2O3 —> V2O5 + 6FeO

Explanation:

The skeletal equation for the reaction is given below below:

VO + Fe2O3 —> V2O5 + FeO

We can balance the equation above by doing the following:

There are 2 atoms of V on the right side and 1 atom on the left side. It can be balance by putting 2 in front of VO as shown below:

2VO + Fe2O3 —> V2O5 + FeO

Now, we have a total of 5 atoms of O on the left and 6 atoms on the right side. We can balance it by putting 3 in front of Fe2O3 and 6 in front of FeO as shown below:

2VO + 3Fe2O3 —> V2O5 + 6FeO

Now, we can see that the equation is balanced

4 0

2 years ago