You're able to figure this out by using the ideal gas equation PV=nRT Where R=.08 (L*atm/mol*K) Convert 37 C to Kelvin=310 K and 102 kPa to atm=roughly 1 atm Manipulate the ideal gas equation to n=PV/RT Plug in your values (1 atm * 6 L) / (.08 (L*atm/mol*K) * 310 K) All units should cancel except for mol and you should get .24 mol. Multiply the molar mass of air by your answer. .24 mol * 29 g/mol moles should cancel and you should get 6.9 g.
Answer:
2 double bonds and 1 single bond
Explanation:
The structural formular of the chlorate ion is given in the attached image.
The bonds present are;
- A single bond between Cl and O
- Two double bonds between Cl and O
<span>C3H7
The unknown compound consists of only carbon and hydrogen. The oxygen comes from the air. So you need to first determine the relative moles of hydrogen and carbon that are present in the CO2 and H2O. First, look up the molar masses of Carbon, Hydrogen, and Oxygen. Then determine the molar masses of CO2 and H2O.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of CO2
1 * 12.0107 + 2 * 15.999 = 44.0087
Molar mass of H2O
2 * 1.00794 + 1 * 15.999 = 18.01488
Number of moles of CO2
22.1 g / 44.0087 g/mol = 0.502173
Number of moles of H2O
10.5 g / 18.01488 g/mol = 0.582852
Since there's 1 carbon atom per CO2 molecule, there are 0.502173 moles of carbon.
Since there's 2 hydrogen atoms per H2O molecule, there are 2 * 0.582852 = 1.165703 moles of hydrogen.
Now we need to find a simple ratio of small integers that comes close to the ratio of 0.502173 / 1.165703 = 0.43079 to determine the empirical formula.
3/7 = 0.428571, an error of only 0.002219. The next closest ratio has an error of 0.013654, over 6 times larger.
So the empirical formula is C3H7</span>
Answer:
The answer to your question is: 14 g of N₂
Explanation:
Data
MW N2 = 14
MW H2 = 1
MW NH3 = 17
N₂(g) + 3H₂(g) ---> 2NH₃(g).
28g 6g 34g
28 g of N₂ -------------------- 6 g of H₂
x -------------------- 3 g of H₂
x = (3 x 28) / 6
x = 14 g of N₂
The answer is C. air moving from low pressure to high pressure.
