Question

chem conversion practice

Answer 1

@zarkam21

how's this for starters?

(6) How many atoms of hydrogen can be found in 45 g of ammonia, NH3?
We will need three unit factors to do this calculation, derived from the following information:

1 mole of NH3 has a mass of 17 grams.
1 mole of NH3 contains 6.02 x 1023 molecules of NH3.
1 molecule of NH3 has 3 atoms of hydrogen in it.

Answer 2

Perfect.

Answer 3

I don't even know where to start >~<

Answer 4

well idk how much chemistry you've learned but how about start by converting 45 g to moles using the given conversion factor?

Answer 5

so 45g* 1 mol NH3/17 grams?

Answer 6

yup

Answer 7

would this be good for the first step because then grams cancel out

45g* 1 mol NH3/17 grams * 1 mol of NH3 / 6.02 x 1023 molecules of NH3

Answer 8

?

Answer 9

Im a little confused on this second part

Answer 10

almost

you sometimes need to flip conversion factors so the units cancel out

Answer 11

but isnt the 6.02 thing cancelling out with the 1 mol NH3 or no

Answer 12

wiat is molecules and mol the same thing

Answer 13

both 6.022 * 10^23 molecules / 1 mole and 1 mole / 6.022 * 10^23 molecules are valid conversion factors between molecules and moles. whichever you use depends on the situation

since we want molecules in the numerator and we want moles to cancel out we use

6.02 x 1023 molecules of NH3 / 1 mole

Answer 14

oh, no, molecules and moles are different

you'll learn this very soon but 1 mole = 6.022 * 10^23 molecules

Answer 15

(technically 1 mole is 6.022 * 10^23 of anything, just like "a dozen" is 12 of anything)

Answer 16

okay and then after that it would be 3 atoms of hydrogen/1 molecule of NH3 so that way NH3 cancels out

Answer 17

perfect

Answer 18

so this is the complete solution (taken from the original website)

Answer 19

as soon as you're ready we can move on to question 2 or stay on this one if you're still confused about anything

Answer 20

I want to know that when you go and solve the equation once you have the numbers all lined up. Do you include the cancelled out numbers

Answer 21

or do only the units cancel out

Answer 22

and how would i multiply just accross?

Answer 23

just the units

so in the above case you would do

6.022e23 * 3 / 17

Answer 24

or if there is a fraction like 1mol of MH3/17 grams.... do i divide first

Answer 25

all the numbers in the top multiply together to get the new numerator

all the numbers in the bottom multiply together to get the new denominator

then just divide new numerator/new denominator and convert to the proper # of sig figs

Answer 26

okay so 45 * (6.022 x 10^23) * 3 = 8.1297 10^25
=17 for the denominator

====4.78 x 10^24

Answer 27

which is really 4.8 x 10^24

Answer 28

perfect

Answer 29

oh okay we can move on i think im understanding now

Answer 30

Q2) How many millimeters are present in 20.0 inches?

1 inch = 2.54 cm
1 cm = 10mm

Answer 31

508?

Answer 32

508mm

Answer 33

perfect

Answer 34

oh thank god

Answer 35

this one is good

The volume of a wooden block is 6.30 in3. This is equivalent to how many cubic centimeters?

remember that the conversion factor

1 inch = 2.54 cm has to be cubed since the dimensions are also cubed

Answer 36

103.23

Answer 37

good just keep sig figs in mind

since 6.30 only has 3 sig figs we'd round to 103 cm^3

Answer 38

well this is the last one on the website I found

A sample of calcium nitrate, Ca(NO3)2, with a formula weight of 164 g/mol, has 5.00 x 1027 atoms of oxygen. How many kilograms of Ca(NO3)2 are present?

Answer 39

okay so I have a question for the cubed questiob

Answer 40

sure

Answer 41

why is it that only the 2.54 is actually to the third power and for the other numbers the cubed sign is just like there and not really changing the vALUE

Answer 42

hm

that's an interesting question

if the conversion factor is like

1 unit^3 = another unit^3

where the cubing is already done, you should just be able to use the conversion factor like normal

but since we are only given 1 inch = 2.54 cm which is just in 1 dimension we have to cube it

Answer 43

drawing it out might give an idea of what happens:

Answer 44

so clearly if we want the volume of this cube we would take either 1 * 1 * 1 cm^3 or 2.54 * 2.54 * 2.54 cm^3

Answer 45

so whether you square the conversion factor or not depends on what you are given and what you are working w/

Answer 46

Oh okay so the number that is like already cubed it just kind of stays as it but obivously the one that needs to be converted has to actually be brought to the third power and solved

Answer 47

yes

Answer 48

I found a worksheet with volume and area conversion if you want more practice with that

Answer 49

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
well this is the last one on the website I found

A sample of calcium nitrate, Ca(NO3)2, with a formula weight of 164 g/mol, has 5.00 x 1027 atoms of oxygen. How many kilograms of Ca(NO3)2 are present?
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 50

oh right

Answer 51

I dont know where to start on this one

Answer 52

they tell you the sample has

5.00 x 10^27 atoms of oxygen so start by converting this to moles

Answer 53

ANd for the worksheet I can like print it out and do it and have yuo just look it over when im done

Answer 54

Does this have to do with it?


1 mole of NH3 has a mass of 17 grams.
1 mole of NH3 contains 6.02 x 1023 molecules of NH3.
1 molecule of NH3 has 3 atoms of hydrogen in it.

Answer 55

HOw do you know that atoms need to be converted to moles

Answer 56

you'll probably learn this later but since they want the mass (in kg)

the general conversion path is

atoms, molecules, or formula units --> moles ---> grams

Answer 57

okay so 16grams/1mole?

Answer 58

I don't actually know how many moles are in oxygen?

Answer 59

hm not quite, remember that molecules and moles aren't the same thing

you'd use 1 mol/6.022*10^23 atoms first

Answer 60

oh okay

Answer 61

then from there you can use 16g/mol to convert to g, then 1kg/1000g to convert to kg

Answer 62

ah, just realized I skipped a step

Answer 63

27 * 10^-26

Answer 64

ohh okay

Answer 65

don't really need to worry about how to figure this out now

but

Ca(NO3)2

has 6 oxygen atoms per mole Ca(NO3)2

so after you divide
5.00 x 10^27 atoms of oxygen by 6.022 * 10^23 you get moles of oxygen atoms

so then you'd divide by 6 to get moles of Ca(NO3)2

Answer 66

then from there you can use 16g/mol to convert to g, then 1kg/1000g to convert to kg

Answer 67

132846.2305

Answer 68

hm that's a bit too much

my diagram is a bit unclear but the #'s on the top are

5e27*16

and the #'s on the bottom are

6.022e23 * 6 * 1000

Answer 69

2.21

Answer 70

like its 8 x10^28 / 3.613

Answer 71

i think you might have missed a decimal point somewhere? :S i end up getting 22.1kg

but the conversion is the important part anyway

Answer 72

5e27*16/(6.022e23*6*1000)

Answer 73

22.2

Answer 74

hm. putting it into my calc gives me

22.141 something so it would round down

Answer 75

to 22.1

Answer 76

YEah i prob am just putting a digit in wrong

Answer 77

http://faculty.chemeketa.edu/tmerzeni/mathcenter/Math060/Mod3/DimAn2.pdf