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postnew [5]
3 years ago
15

Part a use valence bond theory to devise a hybridization and bonding scheme for co2. match the words in the left column to the a

ppropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
4 0
Q1)
first lets draw the lewis structure for CO₂
C is the central atom with O atom on either side of C
O - C - O
the number of valence electrons around each atom 
C- 4
O - 6
O - 6
total number of valence electrons - 16
number of electron pairs - 16/2 = 8
1 pair is shared between 1 O atom and C and another electron pair is shared between C and other O atom 
that leaves us with 8 -2 = 6 pairs 
add 3 pairs each for O atoms as lone pairs and octets for both O atoms are completed. But in C the octet is not complete yet. So we convert the 2 single bonds between C and O to 2 double bonds on either side. then octet of C is complete. number of lone pairs around O is reduced to 2 lone pairs.
the structure is as follows
  ..            . .
: O = C = O :
to determine the geometry lets use VSEPR
number of valence electrons around C - 4
number of bonds around C (2 O with 2 electrons contributed - 4
total number of electrons - 8
number of electron pairs - 4
VSEPR geometry for 4 pairs with no lone pairs is linear geometry

1. The lewis structure for CO₂ has a central carbon atom attached to oxygen atoms through two double bonds.

2. Carbon dioxide has a linear electron geometry

3. Carbon atom is sp hybridized.

4. Carbon dioxide has two <span>C(p)-O(p)</span> π bonds and two O(sp²)-C(sp) sigma bonds.
the geometry of CO₂ linear means its sp hybridised where each orbital has 50% s and 50% p character. One s orbital and one p orbital have mixed and form 2 sp orbitals. the remaining p orbitals take part in π bonds and sp orbitals take part in sigma bonding. So between C and O there's one sigma bond and one pi bond. Altogether 2 sigma bonds and 2 pi bonds.
O has 3 sp² orbitals with 2 of them being lone pair orbitals. One sp² orbital bonds with sp orbital of C forming a sigma bond. the other p orbital of O bonds with p orbital of C forming a pi bond.

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Consider the substances below.
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Answer:

They have the same number of atoms. = YES

They have different masses. = YES

Explanation:

1 mol of beryllium

• 1 mol of salt

beryllium  = Be = Atomic mass: 9.012182

salt  = NaCl = Molar mass: 58.44 g/mol

1 mol of water

• 1 mol of hydrogen

water  = H2O = Molar mass: 18.01528 g/mol

hydrogen = H = 1g/mole

Which statement is true about these substances?

They have exactly the same mass. = NO

They have different numbers of particles = NO

They have the same number of atoms. = YES

They have different masses. = YES

Avogadro constant means the number of units in one mole of any substance (defined as its molecular weight in grams) is equal to 6.02214076 × 10^{23}.

6 0
3 years ago
P+O2 → P4O10
Assoli18 [71]
The answer is B, this is a synthesis reaction. A good way to identify a synthesis reaction is when there are multiple reactants, but only one product.
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3 years ago
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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

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Answer:

As a result, electrolyte solutions readily conduct electricity. ... By contrast, if a compound dissociates to a small extent, the solution will be a weak conductor of electricity; ... Typically, nonelectrolytes are primarily held together by covalent rather than ionic bonds. ... Explain why some molecules do not dissolve in water.

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3 years ago
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