Answer:
mole fraction of N_2 O = 0.330
mole of fraction SF_4 = 0.669
PRESSURE OF N_2 O = 39127.053 Pa
pressure of SF_4 = 792126.36
Total pressure = 118253.413 Pa
Explanation:
Given data:
volume of tank 8 L
Weight of dinitrogen difluoride gas 5.53 g
weight of sulphur hexafluoride gas 17.3 g
Amount of 
amount of 
mole fraction of 
mole of fraction
PV = nRT
P of N_2 O 
mole of SF_4
Total pressure = 39127.053 + 79126.36 = 118253.413 Pa
Mixture is a combination of two or more substances. The components of the mixture are not chemically reacted and can be separated using physical means.
compound is a combination of two or more substances that are chemically reacted or bound together. The components of the compound cannot be separated except by chemical reactions.
Answer:
<span>(4) consist of two or more substances</span>
Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:

- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:

Entonces se obtiene:
a) 72 °F=
=22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F=
= 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g