wicked urgent chemistry questions!! plzzzz answer. showing calculations wud be great?

Question

1. A gas occupies a volume of 0.500L at a pressure of 730 mmHg and 25 degree Celsius. What will be its volume when the pressure is increased by 10% and the temperature is lowered by 5 degree celsius??

2. A mixture of 0.200g of Helium and 0.200g of H2 is confined in a vessel of volume 225 mL at 27 degree celsius. What is the partial pressure of each gas? What is teh total pressure of teh gaseous mixture?

3. 2KClO3(s)-> 2KCl(s)+3O2 (g) Calculate the volume of O2 produced at 25 degress celcius and 630mmHg when 50/0g of KCl)3 undergoes decomposition.

Answer 1

1) The formula is (p1)(v1)/ (t1) = (p2) (v2)/ (t2)

You have all the information you have to plug it in.
p1 = 730 mmHg
v1 = .500 L
t1 = 25 C . You have to convert to Kelvin so you add 273.15 to 25 and get 298.15 K.

p2 = 10% of 730 is 73 so you add it to 730 and get 803 mmHg
v2 = ?
t2 = 25 C + 5 C = 30 C which is (273.15 + 30) and equals 293.15 K.

now plug it into the formula.
(730 mmHg)(.500 L)/298.15 = (803 mmHg) (? L)/ 293.15

cross multiply and get .447 L

2) First convert the grams He and H2 to mol.

.200 g He x (1 mol He/ 4 g He) = 0.05 mol He


.200 g H2 x (1 mol H2/ 2 g H2) = 0.1 mol H2

The formula to find Partial pressure is P= nRT/ V
where n is equal to the mol, R is equal to the constant (0.0821). T is 300.15 because you add 273.15 to 27 C to convert to K. V is 225 mL or .225 L

P (He) = (0.05 mol) (0.0821) (300.15 K)/ (.225 L) = 5.47 atm


P(H2) = (.1 mol) (0.0821) (300.15 K)/ (.225 L) = 10.95

Total pressure is equal to all the partial pressures
5.47 + 10.95 = 16.42 atm

To convert to mmHg you have to multiply atm by 760.

Answer 2

1. P1V1/T1 = P2V2/T2

(730mmHg)(0.500L)/(298K) = (803mmHg)(V2)/(293K)

V2 = (730)(.500)(293)/((298) = 0.360L

2. PV = nRT R = 6.23x10^4 mL-mmHg/mol-K

0.2g He x 1molHe/4gHe = 0.05molHe
0.2gH2 x 1molH2/2gH2 = 0.10molH2, so total moles = 0.15

P(225mL) = (0.15)(6.23x10^4)(300)

P = 12460mmHg total pressure. Partial pressures: P(He) = 12460/3 = 4153mmHg P(H2) = 8307mmHg

3. 2KClO3 ===> 3O2

[50/0g KCl)3? Do you mean 500 g KClO3?)
Atomic weights: K=39.1 Cl=35.5 O=18 KClO3=122.6
500gKClO3 x 1molKClO3/122.6gKClO3 x 3molO2/2molKClO3 x 22.4LO2/1molO2 x 630mmHg/760mmHg x 298K/273K = (500)(3)(22.4)(630)(298)/(122.6)(2)(760(273) = 124L O2

500gKClO3 is given. The next factor comes from the molecular weight of KClO3. The gKClO3 cancel, leaving moles KClO3. The next factor comes from the balanced equation. The moles KClO3 cancel, leaving moles O2. The next factor comes from the molar volume at STP. The moles O2 cancel, leaving L O2. The factor 673/760 accounts for the production of gas at 630mmHg, so the volume mus5t be less than at 760mmHg, so multiply by a number < 1. The factor 298/273 comes in because the gas is warmer than 273K, so the volume is larger, so multiply by a number > 1.

Answer 3

ok, for number 1 I used the combined gas law, V1(P1)/ T1 = V2(P2)/T2.
T1 = 25C + 273 = 298K
T2 = 20C + 273 = 293K
P1 = 730mmHg
730mmHg x 10% or .1 = 73
P2 = 730 + 73 = 803mmHg
Next I set up the problem...
0.5L(730mmHg)/298K= V2 (803mmHg)/ 293K
to get V2 on its own,
V2=.5L(730mmHg)(293K)/298K(803mmHg)
then, you get V2 = .447L

for number 2, you have to use the variant of the Ideal Gas Law, g/V = P(MM)/R(T), where MM is molar mass and R is the constant(I always use the one for atmospheres, .0821(L(atm)/mol(K) )
temp = 27+273 = 300K
V = 225mL/1000 = .225 L
.200 g He
.200 g H2
molar mass He = 4.003g/mol
molar mass H2 = 2.02g/mol
first I'll do the one for pressure of Helium
since we have everything given , I'll just set up the equation already set to solve for pressure
P = (.225 L)(0.0821(L(atm)/mol(K))(300K)/.200gHe(4.003g/molHe) = 6.92 atm He
now I'll do the one for H2
P = (.225 L)(0.0821(L(atm)/mol(K))(300K)/.200gH2(2.02g/molH2) = 13.7atm H2

on to number 3
we have given, temp = 25C + 273 = 298K
P = 630mmHg/760mmHg = 8.303atm
500 g KClO3
first we have to find the number of moles of O2 using the stoichiometric ratios
500g KCLO3 x 1mol KClO3/123.4 g KClO3 x 3 mol O2 / 2 mol KClO3 = 6.08 mol O2
using the number of moles we can use the Ideal Gas Law, PV = nRT to find the volume
already solving for V the equation looks like this
V = 6.08molO2(0.0821(L(atm)/mol(K))(298K)/8.303atm
= 179 L O2

hope that helped

Answer 4

use PV=nRT
p=pressure in atms, v=volume in liters, n=moles,r=.0821, t=temp. in Kelvins
for #1 and #3