Answer:
83ºC
Explanation:
A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.
The reaction of combustion of propane is:
C₃H₈ + 5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol
<em>1 mole of propane release 2222kJ</em>
10.0g of propane (Molar mass: 44.1g/mol).
10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>
If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):
0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.
Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:
504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.
As initial temperature was 20ºC, final temperature will be:
<h2>83ºC</h2>
Answer:
V = 0.2714 mL
Explanation:
Henry's Law:
∴ Kh CO2 = 0.034 M/atm ....The Henry's constant
∴ Partial pressure: P CO2 = 3.7 atm
⇒ <em>C</em> CO2 = P CO2 / Kh
⇒ <em>C</em> CO2 = 3.7 atm / 0.034 M/atm = 108.823 M (mol/L)
∴ molar mass CO2 = 44.01 g/mol
⇒ mol CO2 = (1.3 E3 mg)(g/1000 mg)(mol/44.01 g) = 0.03 mol CO2
volumen solution:
⇒ V sln = (0.03 mol)(L/108.823 mol) = 2.714 E-4 L
⇒ V sln = 0.2714 mL
I’m pretty sure that the answer is current. Hope this helps
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K
