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wlad13 [49]
3 years ago
11

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)  

=4

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 ;)
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A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the proce
Leno4ka [110]

Answer:

The answers to the questions are;

a. The entropy of sublimation for carbon dioxide (the system) is  

134.07 J/Kmol.

b. The entropy of the universe for this reversible process is 376 J/K.

Explanation:

Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at  the temperature where the sublimation is taking place

a. We note that the mass of the solid CO₂ = 389 g

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ in the sculpture = Mass/(Molar mass)

= (389 g)/(44.01 g/mol) = 8.84 Moles

Entropy of sublimation is given by

ΔS_{sublimation} = S_{vapor} - S_{solid} = \frac{\Delta H_{sublimation}}{T}

Where:

ΔH_{sublimation}  = 26.1 KJ/mol

T = Temperature = –78.5°C = ‪194.65‬ K

Therefore the amount of heat required to cause the 389 g of dry ice to sublime =    26.1 KJ/mol  × 8.84 Moles = 230.695 KJ

Therefore the entropy of sublimation = ΔS_{sublimation} = \frac{230.695 KJ}{194.65 K}

= 1.185 KJ/K

= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol

b. The entropy of the universe is given by;

ΔS_{universe} = \Delta S_{system} + ΔS_{surrounding}  

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS_{universe} = \frac{Q}{ T_{system}}  -\frac{Q}{T_{surrounding}}  

                =1.185 KJ/K - -\frac{230.695 KJ}{285.15K} = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

7 0
3 years ago
A sample of copper with a mass of 63.5g contains 6.02 x10^23 atoms calculate the mass of an average copper atom
romanna [79]

Answer:

1.024 x 10⁻²²g

Explanation:

Data Given:

mass of copper = 63.5 g

no. of atoms of copper = 6.02 x10²³ atoms

mass of of an average copper = ?

Solution:

As 6.02 x10²³ atoms have 63.5 g of mass then what will be the mass of atom.

Apply unity formula

                     63.5 g of copper ≅ 6.02 x10²³ atoms of copper

                     mass of copper atom ≅ 1 atom of copper

Do cross multiplication

                     mass of copper atom = 1 atom x 63.5 g / 6.02 x10²³ atoms

                     mass of copper atom = 1.024 x 10⁻²² g

mass of an average copper atom = 1.024 x 10⁻²² g

6 0
4 years ago
For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of
Mamont248 [21]

Answer:

m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Explanation:

Hello,

In this case, the undergoing balanced chemical reaction is:

Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)

Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:

m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Moreover, the percent yield is:

Y=\frac{12.9g}{17.5g} *100\%=73.7\%

Best regards.

6 0
4 years ago
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