All categories

3 years ago

Show the hybridization of co2

Chemistry

1 answer:

garri49 [273]3 years ago

5 0

Answer:

Hybridization of CO2 (Carbon Dioxide)

You might be interested in

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

What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?

leonid [27]

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

7 0

3 years ago

. The speed of a sound in air is 343 m/s. The sound wave has a frequency of 436 HZ and its period is 0.0023 s. What is the wavel

Taya2010 [7]

Answer: 0.79

Explanation:

6 0

3 years ago

How many electrons will be in the valence of non- H atoms when they make covalent bonds?

NNADVOKAT [17]

Answer:

Please one time click here

Mark brainliest

8 0

3 years ago

The atomic number and mass number do not change during this type of radioactive decay.

lutik1710 [3]

In gamma decay, no change in proton number occurs, so the atom does not become a different element

5 0

4 years ago

Show the hybridization of co2​

Show the hybridization of co2