Question

Calculate the volume (in cm3) of 66 g of water at 1 ∘C.

Answer 1

@Vocaloid

Answer 2

oh lordy

anyway since

density = mass/volume then

1000kg/m^3 = (66/1000)kg / V solve for V

Answer 3

1000kg/m^3 = (66/1000)kg / V

66?

Answer 4

hm not quite

since 1000 = (66/1000)/V

we can swap the V and the 1000 to get

V = (66/1000)/1000

of course, that gives you cubic meters so you'd then have to multiply by 10^6 to get it into cm^3

Answer 5

like, as a general rule

if you have A = B/C you can swap A and C and the expression will remain equal

Answer 6

6.6 x 10^-5

Answer 7

hm.

(66/1000)/1000 gives you the units in m^3

multiplying it by 10^6 puts it in cm^3

so actually when you do the math it does end up being 66 whoops making your original sol'n correct

Answer 8

which actually does make sense, as a general rule 1g of water is about 1 cm^3 under reasonable conditions

Answer 9

Calculate the volume (in cm3) of 66 g of water at −1 ∘C.

FOr this it would be the same right just with a -1 degree celsius

Answer 10

oh actually this is very different since it's below the freezing point and water becomes much less dense

looking @ the graph the density at 1 deg. celcius is like 918 kg/m^3 (I remember we had some trouble w/ that last time >>)

Answer 11

and repeating the calculations

918 kg/m^3 = (66/1000)kg / x

then multiply x by 10^6 to get it in cm^3

Answer 12

60588000

Answer 13

so 6.0588 x 10^3 :S

Answer 14

6.06? if rounded

Answer 15

hm not quite

solving

918 kg/m^3 = (66/1000)kg / x

for x gives us about 0.000071895 m^3

then that times 10^6 is about 72g

Answer 16

* 72 cm^3 not g

Answer 17

again that's assuming I read the density correctly (this graph is weird)

Answer 18

wait 65/1000 = .066

and then that times 918 , i got 60.588

Answer 19

what did I do wrong

Answer 20

hm. to isolate

918 kg/m^3 = (66/1000)kg / x

you wouldn't multiply 918 and (66/1000) together

you would actually swap x and 918 to get

x = (66/1000)/918

Answer 21

oh okay I see

Answer 22

What is the percent change in volume when comparing the volume of 66 g of water at 1 ∘C and at −1 ∘C?

Answer 23

okay so 72 and 66

Answer 24

72-66=6 out of 100

Answer 25

.06 = 6%

Answer 26

hm this question isn't worded well but let's assume the 66 is the starting point and the 72 is the ending point

then it becomes

(72-66)/66, then * 100 to convert to a percentage

(since we are comparing from the baseline 66 we divide by 66 not 100)

Answer 27

percent change is generally (final - initial)/(initial) * 100

Answer 28

9.1

Answer 29

9%

Answer 30

yeah good (I think they give you 1 sig fig with the temperature so 9%?)

Answer 31

it showed as wrong : Use the same equation from Part A, and use the volume at 1 ∘C as the initial value, and the volume at −1 ∘C as the final value. Round to one significant digit.

Answer 32

oh it wants the other way around that's cool

Answer 33

(66-72)/72 * 100 then

Answer 34

-8?

Answer 35

hoo boy here's the tricky part b/c some formulas consider percent change to be always positive, some don't

Answer 36

if it shows you the formula, is there an absolute value sign in the numerator?

Answer 37

ok cool just checking, there's no absolute value so try -8 percent

Answer 38

Since the density of ice is less than the density of liquid water, the volume of the same mass of water must be more. You are told to indicate the change when water freezes, so the value should be positive.

Answer 39

Gave me this :S

Answer 40

so just 8 ?

Answer 41

wait what

then 9 should have been right the first time :S

Answer 42

Its not idk

Answer 43

:/

Answer 44

unless the density we used was wrong rendering this whole calculation meaningless :S

Answer 45

Ugh needs to be submitted by 11:59 and I have two more relative questions to this :S

Answer 46

Antarctica contains 26.8 million cubic kilometers (km3) of ice. Assume that the average temperature of this ice during the summer months is -20 ∘C. If all of this were heated to 4 ∘C and melted to form water, what volume of liquid water would form?

Answer 47

We can move to this one. Time is running out so

Answer 48

its useless sticking to one questioln , you know

Answer 49

I got it right. It was 8 :S

Answer 50

d=m/v in this right

Answer 51

uh yeah

Answer 52

we kind of have to estimate here

-20 degrees Celcius the density is about 920kg/m^3
4 degrees celcius the density is about 1000kg/m^3

Answer 53

so uh, if we have 26.8 km^3 of ice then that's 920 * 26.8 = 24656 kg of ice

then converting that back to volume w/ the new density

24656 kg/x = 1000

x = 24.7 m^3

my best attempt

Answer 54

The volume of Antarctic ice is given as millions of cubic kilometers rather than just cubic kilometers.

Answer 55

oh *** I forgot

Answer 56

so use mill prob

Answer 57

well shouldn't the # be the same either way? :S

24.7 km^3?

Answer 58

idk The volume of Antarctic ice is given as millions of cubic kilometers rather than just cubic kilometers.

Answer 59

thats what it says

Answer 60

26.8 km^3 * 10^9 * 920 gives 24656000000000kg

24656000000000kg/x = 1000 kg/(m^3)

gives

24656000000 m^3 which is 24.7 km^3

Answer 61

A 1.00-L sample of water is heated from 1 ∘C to 100 ∘C. What is the volume of the water after it is heated?

Answer 62

density at 1 celc. is like 1000kg/m^3
density at 100 celc is like 958ish kg/m^3

Answer 63

lets see if we can get this done in 4 mins lol

Answer 64

I got the other one i tried 24.7 * 10^7 and that was it

Answer 65

therefore:

1 L = 0.001 cubic meters

so 1L = 0.001 * 1000 = 1 kg
at 100 deg celc

1 kg / x = 958kg/m^3

x = 0.0010438 m^3 ???

Answer 66

what would that be in sig figs "S

Answer 67

they want 3 so 0.00104??

Answer 68

0.01?

Answer 69

0.00104 would have three sig figs

Answer 70

thats wrong ughh

Answer 71

oh *** they want it in L p robably

Answer 72

1.04?

Answer 73

Nvm I got it. Thank you sooo much . I really appreciate it

Answer 74

whew ;;