Answer:
The accurate assertion is "Sample 1 will have a higher percent transmittance of light, %T, than Sample 2."
Explanation:
The Beer-Lambert Law indicates that absorbance correlates directly with the concentration of a solution.
Absorbance = εLc
ε: molar absorptivity coefficient
L: length of the light path
c: concentration
Thus, assuming ε and L remain constant, an increase in concentration results in greater absorbance and reduced transmittance.
Given that sample 1 has a greater CuSO₄ concentration than sample 2, it subsequently exhibits higher absorbance and lower percent transmittance.
Answer:
See below for the explanation.
Explanation:
- Chloroform features three C-Cl bonds that are polar, while methylene chloride contains just two of these polar bonds. Therefore, chloroform is anticipated to be more polar and have a greater dipole moment when compared to methylene chloride.
- Two factors account for the unexpected trend in dipole moments between methylene chloride and chloroform.
- The first factor is the quantity of hyperconjugative hydrogen atoms in each molecule. Hyperconjugation is associated with the empty d-orbital in the chlorine atom. This process enhances the charge separation within a molecule, leading to a higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogens, while chloroform has just one. Thus, methylene chloride is expected to exhibit greater charge separation than chloroform.
- The second factor is the induction of opposing polarity in a C-Cl bond due to another C-Cl bond present within the same molecule. The greater the opposing polarity induction, the less the charge separation, resulting in a lower dipole moment overall.
- Since chloroform has three C-Cl bonds compared to the two in methylene chloride, it experiences greater opposing polarity induction, which leads to its reduced dipole moment.
Explanation:
Here’s the provided information:
Density of gasoline = 0.77 g/ml
Volume of gasoline = 35 L = 35000 ml (since 1 L = 1000 ml)
Density of a substance is defined as its mass divided by its volume.
Mathematically, Density = 
.
<pthus we="" can="" determine="" the="" mass="" of="" specified="" gasoline="" as="" follows:="">
Density =
0.77 g/ml = 
mass = 26950 g.
It is further noted that burning 1 g of gasoline generates 45.0 kJ of energy.
<pconsequently the="" energy="" produced="" by="" burning="" g="" of="" gasoline="" will="" be="" calculated="" as="" follows:="">
= 1212750 kJ.
<pthus we="" conclude="" that="">the energy released by combusting 35 L of gasoline amounts to 1212750 kJ.
</pthus></pconsequently></pthus>
Answer:
Ethane has a greater boiling point.
Explanation:
When examining the Lewis structures, it's crucial to remember that each atom should ideally possess 8 electrons (excluding hydrogen). Both carbons contain 4 valence electrons. Thus, for methane, placing the hydrogens around carbon allows the carbon to achieve an octet of 8 electrons. In the case of ethane, the atoms bond together, and therefore, three hydrogens need to surround each carbon to ensure that they each have 8 electrons.
The primary distinction between methane and ethane is the presence of an extra carbon. The additional carbon in ethane results in an increased surface area for interactions, which necessitates providing more energy to transition the substance from liquid to gas, leading to ethane exhibiting a higher boiling point.
I hope this assists!
To enable a comparison with other experiments, results must be recorded in moles.
number of moles = mass / molecular weight
the number of moles of Mg(H₂PO₄)₂ = 600 / 218 = 2.75 moles
