Chemistry, 13.11.2020 01:00 Frenchfries13

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Answer from: williamsvilletiles

U chile- anyways

Answer from: swordsman

I can't I'm only 14.

Answer from: Quest

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

Answer from: Quest

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

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Chemistry, 22.06.2019 12:30

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Chemistry, 22.06.2019 22:40

Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats.* in a combustion analysis, a 1.054-g sample of maleic acid yields 1.599 g of co_2 and 0.327 g of h2o.* when 0.609 g of maleic acid is dissolved in 25.00 g of glacial acetic acid, the freezing point drops by 0.82 c. maleic acid does not ionize in glacial acetic acid, which has a freezing-point depression constant of k_{\rm f}=3.90~^\circ {\rm c}/m.* in a titration experiment, a 0.4250- g sample of maleic acid is dissolved in water and requires 34.03 ml of 0.2152 m koh for its complete neutralization.* the ph of 0.215 g of maleic acid in 50.00 ml of aqueous solution is found to be 1.80.part adetermine the empirical formula for maleic acid. enter the atoms in order ofc, h, and o.part bdetermine the molecular formula for maleic acid. enter the atoms in order of c,h, and o.express your answer as a chemical formula.part cbased on the titration data, how many ionizable h atoms are in each molecule of maleic acid, c4h4o4? express your answer as an integer.i got the part a and b which is cho and c4h4o4 respectively, can anyone me for part c? the below is what i have so far from other qa'sfinding the empirical formula for maleic acid, needs this info"a combustion analysis a 1.054-g sample of maleic acid yields 1.599g of and 0.327g of ".the rest of your data is used to find molar mass, molecular formula, equivalent mass, # of h+ ions / molecule & katypically this should be: a 1.054-g sample of maleic acid yields 1.599g of co2 and 0.327g of h2osofind mass using molar masses1.599g of co2 @ (12.01 g carbon) / (44.01 g co2) = 0.436 grams of carbon0.327g of h2o @ (2.016 g hydrogen) / (18.01 g/mol h2o) = 0.037 grams of hydrogen1.054-g sample - 0.436 grams of carbon - 0.037 grams of hydrogen = 0.581 grams of oxygennow use molar masses to find moles0.436 grams of carbon @ 12.01 g/mol = 0.0363 mol c0.037 grams of hydrogen @ 1.008 g/mol = 0.0367 mol h0.581 grams of oxygen @ 16.00 g/mol = 0.0363 mol oyour empirical formula ischo and molecular is c4h4o4

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