Ask your own question, for FREE!
Mathematics 7 Online
Karlottalotta:

What is the weight (in grams) of a liquid that exactly fills a 465.0 milliliter container if the density of the liquid is 0.982 grams over milliliter? Round to the nearest hundredth when necessary and only enter numerical values, which can include a decimal point

SmokeyBrown:

If you know the density of the liquid, and you know how much volume it takes up, then finding the weight of that amount of liquid would simply be: weight = density * volume By the way, is this taken off of an online test or something? Not to make an accusation, but the wording of the problem makes me a little bit suspect

Karlottalotta:

im not sure

kittybasil:

Container: \(465.0\) \(mL\) Liquid density: \(0.982\) \(g/mL\) So we have 465.0 milliliters of a liquid with density 0.9872 grams per milliliter. We can resolve this for weight by multiplication or dimensional analysis:\[465.0mL\cdot\frac{0.982g}{mL}\]You can cross-cancel the milliliters out because one is in the numerator and one is in the denominator. This concept is similar to simplifying fractions.

Karlottalotta:

if im honest I suck at math

kittybasil:

So after doing that, we'd have this:\[465.0\cdot0.982g\]This only has one unit now, grams. So if you multiply you should get the final result.

SmokeyBrown:

Nevermind, it's probably just for an online homework assignment. I don't know how classes do it these days. I should have mentioned the units like kittybasil too; in other questions you might have to convert between units, but they happened to work out nicely for this one

Karlottalotta:

thank you both

kittybasil:

Ooh, I hate those. Those are so irritating. But yeah, this one worked out quite well!

kittybasil:

No problem 😊

SmokeyBrown:

:)

Karlottalotta:

I got 2.71

Karlottalotta:

is that right?

kittybasil:

Er... how did you get that?

1 attachment
Karlottalotta:

ion even know

kittybasil:

It's okay, we can try again. Just checking though - you do get everything I explained before right?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!