Chemistry, 29.07.2019 13:00 danuwell

Asample of a compound of mercury and bromine with a mass of 0.389 g was found to contain 0.111 g bromine. its molecular mass was found to be 561. what are its empirical and molecular formulas?

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Hello!

1) To get the empirical formula, we'll need to do the following:

First, we'll need to determine the mass of mercury (Hg) using the following equation:

mHg=mTotal-mBr=0,389 g - 0,111 g = 0.278 g Hg

mHg=mTotal-mBr=0,389 g - 0,111 g = 0.278 g Hg

Now, we calculate the moles of each atom, using atomic masses:

$molesHg=0,278gHg* \frac{1 mol Hg}{200,592gHg}= 0,001386 molHg \\ \\ molesBr=0,111gBr* \frac{1 mol Br}{79,904gBr}= 0,001389 molBr$

To finish, we express the formula using the moles, and dividing each amount by the lower one (0,001386 mol):

$Hg _{ \frac{0,001386}{0,001386} } Br_{\frac{0,001389}{0,001386}} =Hg_1Br_1=HgBr$

2) To calculate the molecular formula, well need to do the following:

First, we calculate the molecular mass of the empirical formula:

$MMHgBr= AM Hg + AM Br= 200,592 g/mol + 79,904 g/mol \\ \\ =280 gHgBr/mol$

Now, we divide the molecular mass of the compound between the molecular mass of the empirical formula to know how many empirical formulas are in the molecular formula:

$\frac{561g/mol}{280g/mol}=2$

To finish, we multiply the coefficients in the empirical formula by the number we just calculated to get the molecular formula:

$Hg_{1*2} Br_{1*2}=Hg_2Br_2$

Have a nice day!

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.

Answer from: Quest

a is the answer to this question

explanation:

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