The answer is C because you must do all the other things first to get to the that’s part hope that helps
Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
Answer:
phosphorus.
Explanation:
The atomic number of phosphorus is 15 so the protons will be 15.
Answer:
Solve the following problems (assuming constant temperature). Assume all numbers are 3 sig figs. 1. A sample of oxygen gas occupies a volume of 250 mL at 740 torr pressure. ... the gas exert if the volume was decreased to 2.00 liters? ... A 175 mL sample of neon had its pressure changed from 75.0 kPa to 150 kPa.
Explanation:
The law is approximately valid for real gases at sufficiently low pressures and high temperatures. The specific number of molecules in one gram-mole of a substance, defined as the molecular weight in grams, is 6.02214076 × 1023, a quantity called Avogadro's number, or the Avogadro constant.
