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Heidenhofer

@bluejay188

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2017-06-23 05:15

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bluejay188

2017-07-04 03:52

RE: #Chemistry Challenge 5

Don't you need to use it to account for the molar ratios of the balanced equation?

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bluejay188

2017-07-04 03:35

RE: #Chemistry Challenge 5

Ah, that's how many Liters are required for 100 km, yet we're going 608 km. SO instead, 200 should be divided by 30.4 making the answer 1.50 x 10^-4 mol∙l-1∙s-1

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bluejay188

2017-07-04 03:18

RE: #Chemistry Challenge 5

9.11 x 10^-4 mol∙l-1∙s-1 Ok, I think I've figured it out. So the Rate equation for this would be Rate = ((1/2)(d[C8H18]/dt)). You can find d[C8H18] by dimensional analysis from 70.3 kg CO2... which is

bluejay188

2017-07-03 21:10

RE: #Chemistry Challenge 5

1.) 2 C8H18 + 25 O2 ===> 16 CO2 + 18 H2O 2.) 608 km (Dimensional Analysis) 3.) 70.3 kg CO2 3.9 x 10^4 L CO2 (Dimensional Analysis / Ideal Gas Law) 4.) .306 [C8H18]/hr (dc/dt)

bluejay188

2017-07-03 20:27

RE: | KatiepelC | Paramore - Hard Times (my live cover for steemit:)

That was awesome, upvoted!

bluejay188

2017-07-03 16:15

RE: Hey everyone - Piano / I take requests!

bluejay188

2017-07-03 16:13

RE: Hey everyone - Piano / I take requests!

Thanks, and awesome - will do!