please help i have to finish this today!

Question

" Solve the problem; show your work in the space provided and circle the answer. Express the answer using proper significant figures and units."

1. A sample of nitrogen occupies 1.00 L at a temperature of 250. K. What volume would it occupy at 1500. K.?
2. A typical weather balloon contains approximately 563 L of hydrogen gas when it is released. If the temperature and pressure at the surface are 288 K and 1.00 atm when the balloon is released, what will its volume be at 32.0 km above the surface, where temperature and pressure are 299 K and 8.57x10-3 atm?
3. Sodium metal and chlorine gas can be produced industrially by passing an electric current thorugh molten sodium chloride:

2 NaCl(1) → 2 Na(1) + Cl2(g)

What volume of Cl2 at 293 K (20.°C) and 1.00 atm can be obtained from 68.0 g NaCl using the reaction? (MNaCl = 58.44 g/mol; R = 0.0821 L•atm/[K•mol])

please show work for all the questions , thank you

Answer 1

1). Using one of the major ideal gas laws will help you with this one. The one that pertains and that is relevant relating both temperature and volume is Charle's law. Charle's law says:. T1*V1=T2*V2.

Kelvin units help so you don't have to deal with negative volumes and temperatures.

So:. 250*1 = 1500*V2

250/1500= V2

V2 = 0.1666 L. Or 0.167 L

2) You can use this ideal gas law and solve for moles of hydrogen and because you have the same substance ( Hydrogen) plug these moles back into new scenario and solve for second volume:

PV= nRt

1*563 = n(0.0821)288

n = 23.81

Now use this to solve for V or volume, the second volume

(8.57x10^- 3)V = (23.81)(0.0821)299

When you solve for V you get:. 68,203 L or 68,200 L for 3 sig figs. In scientific notation that is, 6.82x 10^4 L

Another way of solving this is by using Boyle's law which is:

P1*V1 = P2*V2. This relates pressure and volume. When you do so, you get roughly the same answer at:

1(563) = (8.57x 10^-3)V2

V2 = 65,694 L or 6.57x10 ^4.

This helps you check your answer. It seems that both methods bring the answer within approximation of 6.70 x10^4 or so in L so you can see that both methods are displaying a large volume in L, liters due to the very small pressure.

3). This is asking you to solve for volume as you are given everything else. From the grams of the NaCl, you can get moles of Cl2 and then volume.

68.0 g of NaCl. *1mol/58.44g. grams cancel top to bottom and you have:

1.163587 mol of NaCl now use balanced equation to find moles of Cl2.

According to the balanced equation:

2mol NaCl ==> 1mol Cl2

1.163587 mol NaCl==> Xmol Cl2

Cross multiply and solve for X:

2X =1*1.163587

X= 1.163587/2

X= 0.58179. mol of Cl2

So now with this, plug in all your knowns and solve for volume of Cl2.

PV =nRt

1*V= 0.58179(0.0821)*293

V = 13.99 L or 14.0 L for 3 sig figs.