How do you correctly round this problem #(30.00 g)/ (10.0 mL)#?

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Answer 1 · 2017-12-30T08:22:14

#(3 g)/(mL)#

You can just divide to get #(3 g)/(mL)#. Is this talking about the density of a substance?

Answer 2 · 2017-12-30T08:29:44

Please see the step process below...

Firstly, you have to convert the #mL -> L#

Though it depends on what you want your final unit to be..

But in this case, am going to convert it;

#mL -> L#

#1000mL -> 1L#

#10mL -> xL#

#x = (1L xx 10mL)/(1000mL)#

#x = (1L xx cancel(10mL))/(100 xx cancel(10mL)#

#x = 1/10L#

#x = 0.01L#

Hence we have;

#(30.00g)/(10.0mL) rArr (30g)/(0.01L)#

#rArr 3000.00gL^-1 or 3.00gmL^-1#

Answer 3 · 2017-12-30T08:57:02

Well, you are dividing, so you keep the fewest number of significant figures. #"10.0 mL"# has #3#, and #"30.00 g"# has four.

All trailing zeroes after the decimal place are significant, and zeroes sandwiched between a decimal place and a nonzero digit are also significant.

So you just get:

#"30.00 g"/"10.0 mL" = color(blue)("3.00 g"/"mL")#