F. hold on to their protons more strongly
Explanation:
When conducting a melting point experiment, if we were to heat a sample quickly. Large amount heat is provided instantly which would melt the crystals in the tube very quickly, even before the temperature of the thermometer reaches to that level. So the observes melting point would be much lower than the actual melting point when sample is heated slowly.
Larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
<h2>Cause of stronger dispersion force
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Larger atoms and molecules have stronger dispersion forces than smaller atoms and molecules because in a larger atom or molecule, the valence electrons are present farther from the nuclei than in a smaller molecule.
They are less tightly held to the nuclear charge present in the nucleus and can easily form temporary dipoles so we can conclude that larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
Learn more about dispersion force here: brainly.com/question/1455074
Learn more: brainly.com/question/26140191
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated
Here are attached photos, I don’t think they ar links, lmk if they don’t work
in the answering section it didn’t show me the full question so I only saw a little of question 5. Hopefully you get the idea of how to identify which equations to use tho! Good luck!
