All categories

3 years ago

How many hybrid orbitals are found in CCl4?

Chemistry

2 answers:

Ipatiy [6.2K]3 years ago

7 0

CCl4

For central atom C

The formula to calculate hybridization is:

n= 1/2 [group no. central atom +no. of monovalent atoms attached to central metal atom (except oxygen) + magnitude of negative charge - magnitude of positive charge]

n= 1/2 [4 +4 + 0-0]

n= 1/2 [8)

The hybridization is sp3 and the electron geometry is tetrahedral

Tatiana [17]3 years ago

3 0

The answer is FOUR 4
hope this helps ;)

You might be interested in

The answer is 51 ml but can someone explain why please

makvit [3.9K]

Answer:

Ok so I dont have a paper on me, but to solve this, first you need to know, 1 ml is 1 cubic centimeter, okay so we need to find the volume, which is 4/3 times pie times r cubed, the answer to this equation will then bet 51 cubic centimeters which directly translate to 51 ml pushed out of the full cup

6 0

3 years ago

Read 2 more answers

Calculate the number of grams of solute needed to make each of the following solutions:

AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

1) 100 g of 0.500% (w/w) NaI

Mass of solution = 100 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{100 g}\times 100$

$x=\frac{0.500\times 100 g}{100}=0.500 g$

0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.

2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{250 g}\times 100$

$x=\frac{0.500\times 250 g}{100}=1.25 g$

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

$1.25\%=\frac{x}{500 g}\times 100$

$x=\frac{1.25\times 500 g}{100}=6.25 g$

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

$2.00\%=\frac{x}{750 g}\times 100$

$x=\frac{2.00\times 750 g}{100}=15.0 g$

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

3 0

3 years ago

Which step in his experiment is the most important if he wants his experiment to correctly demonstrate the conservation of mass?

Oksi-84 [34.3K]

D I’m pretty sure if I’m wrong I’m sorry!!!

7 0

3 years ago

Calculate The molar mass of aluminum nitrate Al(NO3)3

sleet_krkn [62]

Answer:

Molar mass: 212.996 g/mol

Explanation:

Formula: Al(NO3)3

Molar mass: 212.996 g/mol

IUPAC ID: Aluminium nitrate

Melting point: 72.8 °C

Boiling point: 135 °C

Density: 1.72 g/cm³

Soluble in: Water, Alcohol

7 0

4 years ago

Please help with chem final need answers within one hour i’ll give brainliest

Dahasolnce [82]

Answer:

1. 7

2. 7

3. 16.60

Explanation:

1. 7

3x-3=18

+3 +3

3x=21

/3 /3

x=7

2. 7

4x-5=23

+5. +5

4x=28

/4. /4

x=7

3. 16.60

it reaches above the halfway point after 16, and it's one above it. So, it's 16.60

7 0

3 years ago