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sleet_krkn [62]
3 years ago
8

Combined gas law, PLEASE HELP!

Chemistry
1 answer:
katrin2010 [14]3 years ago
5 0
You correctly said that you need the combined gas law. 

combine gas law----> P1V1/T1= P2V2/T2

P1= ?
V1= 2.58 L
T1= 368K

P2= 777 torr
V2= 1.53 L
T2= 495K

Now we plug in the values. 

(P1 X 2.58)/ 368= (777 X 1.53)/ 495 K

P1= 343 torr
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Kc= [3.1*10^-2]^2. (mol/L^2) divide by [8.5*10^-1] [3.1*10^-3]^3 (mol/L)^4​
jok3333 [9.3K]

Answer:

Kc = 12.58

Explanation:

Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3

Kc = (0.052441)(0.10513)/(0.002385)(0.18373)

Kc = 0.0005513/0.000438

Kc = 12.58

Hope that helps!!

5 0
3 years ago
Which process is an example of a chemical change?
WARRIOR [948]

Answer:

A is the closest thing. You change the composition of the steak. You don't in any of the others.

Explanation:

Usually when you cook something, you are doing something to the composition of the object being cooked. A steak might not be obvious, but boiling an egg should be.

Chopping a tree is something physical. You are removing mass in such a way that the tree will fall. There's nothing chemical about that.

Heating a cup of tea looks like it might be chemical. After all steam is sometimes given off which looks like it is chemical. It's not. The water in the tea is just changing phase.

Drying clothes in a dryer. Again, this looks like something might have changed. After all the mass of the clothes just became less. But all you are doing is separating two masses (leaving one of them behind).

8 0
3 years ago
Read 2 more answers
The elements beyond uranium on the periodic table of the elements are called the ------- elements
iris [78.8K]
The answer is elements
4 0
3 years ago
Help I need to finish my chemistry report.
Alexeev081 [22]
The thing that they have in common is that they are all non metals
4 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
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