So to get the answer use the equation: moles = mass/ RAM the RAM is the relative atomic mass and can be calculated by using the bottom number in the box of the element on the periodic table so for this question the mass of P4 is 161g and the RAM of P4 is 124 (31 X 4) so 161/124 = 1.3 and because you need to work out P2O5 and not P4 2 is half of 4 so you half the moles so the answer should be 0.65
Answer:
-238.54 kJ/mol.
Explanation:
- We need to calculate the standard enthalpy of formation of liquid methanol, CH₃OH(l) that has the equation:
C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ?
?? kJ/mol.
- using the information of the three equations:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
(2) H2(g) + ½ O₂(g) → H₂O(l), ΔHf₂° = -285.8 kJ/mol
.
(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃° = -726.56 kJ/mol
.
- We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation:
- equation (1) be as it is:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
- equation (2) should be multiplied by (2) and also the value of ΔHf₂°:
(2) 2H2(g) + O₂(g) → 2H₂O(l), ΔHf₂° = 2x(-285.8 kJ/mol
).
- equation (3) should be reversed and also the value of ΔH₃° should be multiplied by (-1):
(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g), ΔH₃° = 726.56 kJ/mol
.
- By summing the modified equations, we can get the needed equation and so:
The standard enthalpy of formation of liquid methanol, CH₃OH(l) = ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol
) + 2(-285.8 kJ/mol
) - (- 726.56 kJ/mol) = -238.54 kJ/mol.
Answer:
The pressure of CO2 = 0.48 atm
Explanation:
Step 1: Data given
Kp = 0.23
Step 2: The balanced equation
2NaHCO3(s) ↔ Na2CO3(s) + CO2(g) + H2O(g)
Step 3: Calculate the pressure of CO2
Kp = (p(CO2))*(p(H2O))
For 1 mol CO2 we have 1 mol H2O
x = p(CO2) = p(H2O)
Kp = 0.23 = x*x
x = √0.23
x = 0.48
pCO2 = x atm = 0.48 atm
The pressure of CO2 = 0.48 atm
Try 0.80 grams NaOH . M=mol/L if you rework it to get mol, you have mol=0.20M×0.100L which is 0.02mol of NaOH, multiply that by the amu of NaOH (39.998) and should get 0.799g