There are 2 possible answers here : b and d.
The Ideal Gas Equation is : <u>PV = nRT</u>
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Here, when pressure is increased and temperature is lowered, the volume of the molecules will substantially decrease, which means it has deviated from ideal behavior.
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Answer:
Mass = 179.9 g
Explanation:
Given data:
Volume of solution = 450 mL
Molarity of solution = 2.00 M
Mass in gram required = ?
Solution:
Volume of solution = 450 mL× 1 L / 1000 mL = 0.45 L
Molarity = number of moles of solute/ Volume of solution in L
2.00 M = number of moles of solute / 0.45 L
Number of moles of solute = 2.00 M × 0.45 L
M = mol/L
number of moles of solute = 0.9 mol
Mass of CaBr₂ in gram:
Mass = number of moles × molar mass
Mass = 0.9 mol ×199.89 g/mol
Mass = 179.9 g
<h3>Answer:</h3><h2>Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.</h2>
