A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
27g x 1cm^2/2.7g = answer in cm^2
Grams cancel out
It would 3 because if you read it tells you the answer
Answer:
14. 13.2cg = 1.32dg
15. 3.8m = 0.0038km
16. 24.8L = 24800mL
17. 0.87kL = 870L
18. 26.01cm = 0.0002601km
19. 0.001hm = 10cm
Explanation:
14. 13.2/10 = 1.32
15. 38/1000 = 0.0038
16. 24.8(1000) = 24,800
17. 0.87(1000) = 870
18. 26.01/100000 = 0.0002601
19. 0.001hm(10000) = 10
An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.