Answer:
A) An Alkaline
Explanation:
An alkane that is pentane which is substituted by a methyl group at position 3. It is used as a solvent in organic synthesis, as a lubricant and as a raw material for producing carbon black.
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Answer:
Explanation:
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In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:
Thus, for the initial pressure and temperature in kelvins the final temperature in kelvins, we compute the final pressure as:
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Answer:
The value of the carbon bond angles are 109.5 °
Explanation:
CH3CH2CH2OH = propanol . This is an alcohol.
All bonds here are single bonds.
Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.
Sp3- types have angles of 109.5 ° between the carbon - atoms.
This means that the value of the carbon bond angles are 109.5 °
Dryness and temperature are the primary reasons but also the type of honey.
