Answer:
See explanation
Explanation:
Many organic compounds have low melting points. This is due to the fact that many of these compounds are non polar.
However, compound X is slightly polar but still has a melting point which is far less than that of sand composed of a high melting point inorganic material.
Since sand has a much higher melting point compared to compound X, the researcher need not be worried that sand was spilled into his beaker.
Answer:
True => ΔH°f for C₆H₆ = 49 Kj/mole
Explanation:
See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.
Due to hydrogen bonding there is a formation of cage like structure called lattice in ice due to which <span> density of ice is less than that of water. Moreover, it is a known fact that density of water is maximum at 4°C.</span>
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
