Answer:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
Explanation:
The word equation for the combustion of propane can be obtained from the chemical equation;
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The word equation is therefore:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
For such a combustion reaction, carbon dioxide and water are produced in the process.
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1
number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂
we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation
101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL
Answer:
Explanation:
DNA polymerase is an enzyme that synthesizes DNA molecules from deoxyribonucleotides, the building blocks of DNA. These enzymes are essential for DNA replication and usually work in pairs to create two identical DNA strands from a single original DNA molecule.
Answer: 1.48 atmosphere
Explanation:
Pressure in kilopascal = 150
Pressure in atmosphere = ?
Recall that 1 atmosphere = 101.325 kilopascal
Hence, 1 atm = 101.325 kPa
Z atm = 150 kPa
To get the value of Z, cross multiply
150 kPa x 1 atm = 101.325 kPa x Z
150 kPa•atm = 101.325 kPa•Z
Divide both sides by 101.325 kPa
150 kPa•atm/101.325 kPa = 101.325 kPa•Z/101.325 kPa
1.48 atm = Z
Thus, 150 kPa is equivalent to 1.48 atmospheres
Answer:
The standard enthalpy of formation of this isomer of 
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
![\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)
Substitute the all values in the entropy change expression.
![-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
