2. This is like a mountain but smaller. A. valley B. hill C. plateau D. plain

Use the oxidation rule to determine the oxidation number of carbon in H2CO3

torisob [31]

In sure u will do great and the answer will be right

3 0

3 years ago

If solid iron is dropped in liquid iron, it will most likely?

S_A_V [24]

it will sink

Explanation:

the solid iron will sink because it is dense than the liquid iron I will sink and it will melt

4 0

3 years ago

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Is this equation completely balanced? 2C8H8 + 25O2 8CO2 + 18H2O No, because the number of carbon, hydrogen & oxygen atoms on

Sergio039 [100]

So to make it clear let's break the equation down species by species and assess the number of each species on bothe sides of the equation:
2C₈H₈ + 25O₂ → 8CO₂ + 18H₂O
LHS: C - 16 RHS: C - 8
H - 16 H - 36
O - 50 O - 34

Thus based on that it is evident that the equation is not quite balanced. This therefore means a "No, because the number of carbon, hydrogen & oxygen atoms on both sides of the equation are not equal."

The actual balance equation would be C₈H₈ + 10O₂ → 8CO₂ + 4H₂O

8 0

3 years ago

Using the black numbers on the stopwatch to answer the questions.

Rus_ich [418]

Answer: the top one is 5.3 s and the bottom one is tenths of seconds

Explanation:

4 0

4 years ago

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

tatiyna

Answer:

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%
The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

4 0

3 years ago