Question

Help!

Answer 1

this is more confusing than what i am doing

Answer 2

What is the density of a piece of metal that
has a mass of 8.94 × 10^5 g and is 95.5 cm × 47.4 cm × 98.3 cm?
Answer in units of g/cm3

I got this far 8.94 x 10^5/44497.61 but not sure how to simplify it to the answer

Answer 3

I would just go about using 6ft as height in the volume formula and just say that greater than or equal to this much glucose would have to be fermented to achieve CO2 level that is 6ft or higher

Or you can just use 2 mole of CO2, no balancing required.

volume of a cube = Length*height*width


also note, 1 liter = 1000cm^3

not any of it, you just use the ratio convert Liters to cm^3 in the ratio L/mole then multiply moles of CO2 by that ratio to find volume that CO2 takes up

then just plug and chug into the formula for the volume of a cube

Volume that CO2 takes up in cm^3 = Length of Cellar(cm)*x*Width of Cellar(cm)

x = height of CO2 gas

Alternatively you could just use x = 6ft (convert 6ft to cm) and solve for volume then work backwards to figure out the amount of moles of glucose that needs to be consumed if you want to see the minium amount of glucose that needed to be fermented to reach 6ft.

This question is just a lot of conversions

https://www.youtube.com/watch?v=8jB-LaTGgq8

Answer 4

You were right, I am very sorry! The glucose was given. There was a previous question where i have found the mass of glucose in grams, and now the question makes more sense to me. Thanks for clarifying it for me @Australopithecus and thanks for the input @srossholmes. Again, i apologize for giving you guys a hard time yesterday!

Answer 5

I knew I was right, but I'm glad I was helpful and that you figured it out