SOLUTION: A new lab technician, on his first day on the job, drops his lab key into a large bottle of concentrated nitric acid, and key begins to dissolve. the mass of the key is 2.03 oz (1

Question 616946: A new lab technician, on his first day on the job, drops his lab key into a large bottle of concentrated nitric acid, and key begins to dissolve. the mass of the key is 2.03 oz (1 oz=28.35g) and has a density of 6.347 g/cm^3. The acid dissolves the key at a rate of 0.00235 g/s. By the time the lab tech gets the key out of acid, 10.0 minutes have passed.What is the volume of the key by then?
I solved this AMP Parsing Error of [10min*(60sec/1min)*(0.00235g/1 sec)*(1oz/28.35g)*(1g/2.03oz) = 0.025g]: Invalid function '/1\min)*(0.00235\g/1sec)*(1\oz/28.35\g)*(1\g/2.03\oz)=0.025\g': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .

then but I feel like the answer is wrong.
Found 2 solutions by ewatrrr, solver91311:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi,
If I am understanding correctly:
2.03 oz originally, (1 oz=28.35g)
I. originally
II. AMP Parsing Error of [ 10min * (60sec/1min) * (0.00235g/1 sec)]: Invalid function '/1\min)*(0.00235\g/1sec)': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. . = 1.41gm Grams dissolved
III. 57.55 - 1.41 = 56.14gms left after 10 min
What is the volume of the key by then?
V = mass/density

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

First calculate the number of grams of mass in the key before it went in the acid.

Next, calculate the number of grams lost in 10 minutes:

Next, calculate the mass remaining after 10 minutes, i.e. what it was minus what it lost:

Then divide the final mass by the density to calculate the volume:

John

My calculator said it, I believe it, that settles it

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