Answer:
17.27 atm
Explanation:
Use the ideal gas law or PV = nRT
We are solving for pressure here so lets isolate for P before we plug in values:
![P = \frac{nRT}{V}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BnRT%7D%7BV%7D)
So first to get n or the number of moles we need to convert the grams of N2O to moles of N2O. We can do this by multiplying by the inverse of the molar mass like so:
![\frac{519.93g(N2O)}{1} \frac{mol}{44.013g(N2O)}](https://tex.z-dn.net/?f=%5Cfrac%7B519.93g%28N2O%29%7D%7B1%7D%20%5Cfrac%7Bmol%7D%7B44.013g%28N2O%29%7D)
Our grams of N2O would cancel and give us 11.813 mol of N2O
Now all thats left is to plug in and solve with the correct value for R which in this case for all of our units to cancel is 0.08206
![P = \frac{(11.813)(0.08206)(125.63)}{7.05}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%2811.813%29%280.08206%29%28125.63%29%7D%7B7.05%7D)
P = 17.27 atm
(I would double check the calculator work if it is for correctness just be sure)
They burn up and cause it to look like shooting stars
Ar (Cl) = 34.969 * 75.78/ 100 + 36.966 * 24.22/ 100= 35, 45 :)
<span>Water is nonpolar covalent, so all substances dissolve in it. </span>
Answer:
50.00 g of NO
Explanation:
Remember that the balanced chemical reaction equation is indispensable in solving any question that has to do with stoichiometry. Hence the first step in solving the problem is noting down the balanced chemical reaction equation.
2NO(g) + O2 (g)→ 2NO2(g)
Now we try to find out the reactant in excess. The reactant in excess gives the greater mass of product.
For O2;
From the balanced reaction equation;
32 g of O2 yields 92g of NO2
16.00g of O2 will yield 16.00×92/32 = 46g of NO2
For NO;
30g of NO yields 92g of NO2
80.00 g of NO yields 80.00 × 92/30 = 245.33 g of NO2
Hence NO is the reactant in excess.
If 1 mole of O2 reacts with 2 moles of NO2 according to the balanced reaction equation
Then 32 g of O2 reacts with 60g of NO according to the balanced reaction equation
16.00 g of O2 reacts with 16.00 × 60 /32 = 30 g of NO
Hence mass of excess reactant used in the reaction = total mass of NO- mass of NO reacted= 80.00g -30.00g = 50.00 g of NO
Hence the mass of excess reactant used in the reaction is 50.00 g of NO