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sasho [114]
3 years ago
7

The volume of a gas is 36.0 ml at 10.0°c and 4.50 atm. at what temperature (°c) will the gas have a pressure of 3.50 atm and a v

olume of 85.0 ml?
Chemistry
1 answer:
galben [10]3 years ago
5 0
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
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Can anyone help with thiss??
IrinaVladis [17]
  1. l
  2. kg
  3. cm
  4. gm
  5. ml
  6. km,m
  7. m
  8. nm
  9. mg
  10. ms
  11. gm
  12. nm
  13. .
  14. .
  15. l,gm
  16. .
  17. kg
  18. m
  19. gm,m
  20. s

Hope this helps :-). And....

ml: milliliter

kg: kilogram

m: meter

mg: milligram

mm: millimeter

ms: millisecond

l: liter

km: kilometer

µg: micro gram

cm: centimeter

g: gram

nm: nano meter


5 0
3 years ago
Please help me!!!!!!!!
Oksanka [162]
Your answer would be surface runoff 
5 0
3 years ago
At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2
s344n2d4d5 [400]
Answer is:  0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K. 
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0
3 years ago
F.ree points for all!
shepuryov [24]

Answer:

here...I helped u! XD

jkjk...I'll got to the other questions too...dw! XD

Explanation:

4 0
3 years ago
How much heat is required to raise the temperature of 81.0 g of water from its melting point to its boiling point?
Dovator [93]

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

mass of water = 81 gram

Initial temperature = 0°C

Final temperature = 100°C

Specific heat of water = 4.184

Find:

Required heat Q

Computation:

Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

Q = (81)(4.184)(100-0)

Q = 33,890.4

Specific heat of water = 33.89 KJ

6 0
3 years ago
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