Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
Answer:
2 H⁺ + 2e = H₂ ( reduction )
Explanation:
Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂
Fe( s ) = Fe⁺² + 2e ( oxidation )
2 H⁺ + 2e = H₂ ( reduction )
moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
[H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is [H2SO4] = 0.07729 M
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
It's called Ice Wedging :)
