Chemistry, 02.09.2019 02:30, adkinsryan72

15 g of hydrogen reacts with 25 g of nitrogen. how much excess reagent will be left over when the reaction stops?
5.36 g of excess
7.23 g of excess
12.12 g of excess
19.27 g of excess

15 g of hydrogen reacts with 25 g of nitrogen. how much excess reagent will be left ove

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answered: ronaldhernandez598

1) Chemical reaction: 3H2 + N2 -> 2NH3

2) proportions 3 moles H2 ; 1 mol N2 : 2 mol NH3

3) molar masses

H2: 2*1g/mol = 2 g/mol

N2: 2* 14 g/mol = 28 g/mol

NH3: 14g/mol + 3 g/mol = 17 g/mol

4) Conversion of the data in grams to mol

#mol = mass / molar mass

# mol H2 = 15 g / 2 g/mol = 7.5 mol

# mol N2 = 25 g / 28 g/mol = 0.89 mol

5) Compare the ratio of the reagents with the theoretical ratio:

Reagents: 0.89 mol N2 : 7.5 mol H2

Theoretical: 1mol N2 : 3 mol H2

You conclude that the H2 is in excess and need to calculate how many moles react with 0.89 mol of N2.

6) moles of N2 that react:

x mol H2 / 0.89 mol N2 = 3 mol H2 / 1 mol N2 => x = 0.89*3/1 = 2.67 mol H2

7) Calculate the excess as Initial # mol H2 - # mol that reacted = 7.5 mol - 2.67 mol = 4.83 mol H2 in excess

8) Convert mol of H2 to grams of H2

mass = # mol * molar mass

mass = 4.83 mol * 2 g/mol = 9.66 grams in excess

answered: Guest

c. because step up transformer steps up the input voltage i.e increases the input voltage

the number of turns are greater on the secondary side of step up transformer

transformer does not handle dc current

answered: Guest

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.