When the Lewis structure for HCOOCH3 (methyl methanoate) is drawn, how many bonding domains and how many non-bonding domains (lo
ne pairs of electrons) are present?
1 answer:
Answer : There are 16 number of bonding electrons and 8 number of non-bonding electrons or lone pairs of electrons.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 2(4) + 4(1) + 2(6) = 24
According to Lewis-dot structure, there are 16 number of bonding electrons and 8 number of non-bonding electrons or lone pairs of electrons.
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Answer:
1. Molarity of MgSO₄ = 0.6 M
2. Molarity of AgNO₃ = 0.06 M
3. volume of acetic acid = 250 mL
4. Molarity of NaCl= 2.3 M
5. Mass of C₁₂H₂₂O₁₁ = 4.1 g
6. Mass of C₁₀H₈ (naphthalene) = 77 g
7. mass of ethanol = 11 g
8. Mass of DDT = 0.011 mg
Explanation:
Ans 1.
Data given
mass of MgSO₄ = 403
Volume of solution = 5.25 L
Solution:
First we will find mole of solute
no. of moles = mass in grams / Molar mass . . . . . .(1)
Molar mass of MgSO₄ = 24 + 32 + 4(16)
Molar mass of MgSO₄ = 120 g/mol
Put values in equation 1
no. of moles = 403 g / 120 g/mol
no. of moles = 3.36 mol
Now to find molarity
Formula used
Molarity = No. of moles of solute / solution in L . . . . . . (2)
Put Values in above equation
M = 3.36 mol / 5.25 L
M = 0.64
So, round the figure to two significant digits.
Molarity of MgSO₄ = 0.64 M
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Ans 2.
Data given
mass of AgNO₃ = 1.24 g
Volume of solution = 125 mL
convert mL to L
1000 mL = 1 L
125 mL = 125/1000 = 0.125
Solution:
First we will find mole of solute
no. of moles = mass in grams / Molar mass . . . . . .(1)
Molar mass of AgNO₃ = 108 + 14 + 3(16)
Molar mass of AgNO₃ = 170 g/mol
Put values in equation 1
no. of moles = 1.24 g / 170 g/mol
no. of moles = 0.0073 mol
Now to find molarity
Formula used
Molarity = No. of moles of solute / solution in L . . . . . . (2)
Put Values in above equation
M = 0.0073 mol / 0.125 L
M = 0.06
So, round the figure to two significant digits.
Molarity of AgNO₃ = 0.06 M
______________________
Ans 3.
Data Given:
volume of acetic acid = 500 mL
% solution of acetic acid (M)= 50%
volume of acetic acid needed = ?
Solution:
formula used
percent of solution = volume of solute/ volume of solution x 100
Put Values in above formula
50 % = volume of solute / 500 mL x 100
Rearrange the above equation
volume of solute = 50 x 500 mL /100
volume of solute = 250 mL
So, volume of acetic acid = 250 mL
______________________
Ans 4
Data Given:
Molarity of NaCl (M1) = 6 M
Volume of NaCl (V1) = 750 mL
convert mL to L
1000 ml = 1 L
750 ml = 750/1000 = 0.75 L
Volume of dilute solution (V2)= 2 L
Molarity of dilute solution (M2) = ?
Solution:
Dilution Formula will be used
M1V1 = M2V2 . . . . . . (1)
Put values in equation 1
6 M x 0.75 L = M2 x 2 L
Rearrange the above equation
M2 = 6 M x 0.75 L / 2 L
M2 = 2.25 M
So, round the figure to two significant digits.
Molarity of NaCl= 2.3 M
_________________________
Ans 5.
Data Given:
Molarity of C₁₂H₂₂O₁₁ = 0.16 M
Mass of C₁₂H₂₂O₁₁ (m) = ?
Volume of dilute solution = 75 mL
convert mL to L
1000 ml = 1 L
75 ml = 75/1000 = 0.075 L
Solution:
As we know
Molarity = no.of moles/ liter of solution . . . . . . .(1)
we also know that
no.of moles = mass in grams / molar mass . . . . . (2)
Combine both equation 1 and 2
Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)
Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)
Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176
Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol
Put values in equation in equation 3
0.16 M = (mass in grams / 342 g/mol) / 0.075 L
Rearrange the above equation
mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol
mass in grams = 4.104 g
So, round the figure to two significant digits.
Mass of C₁₂H₂₂O₁₁ = 4.1 g
____________________
The remaing portion is in attachment
