Answer:
a) 2.173
b) 20ml
c) 3.87
d) 4.35
e) 8.8
f) 12.45
Explanation:
chech the attachment for explanationns
Answer : The correct option is, pressure.
Explanation :
The ideal gas equation is,

where,
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
T = temperature of the gas
R = gas constant
The value of 'R' has several different values which are :




That means, the value of 'R' is different due the change in the pressure value and all the variables (temperature, volume and moles) are constant.
Hence, the correct option is, pressure.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
The magnitude of a star as it would appear to a hypothetical observer at a distance of 10 parsecs or 32.6 light-years. This rates how visible celestial bodies are when they are all viewed from the same distance. Luminosity: The brightness of a star in comparison with that of the sun.
Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL
Explanation:
The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.
<em>Number of moles of oxygen gas = mass/molar mass</em>
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = density * volume
mass of oxygen gas = 1.149 g/ml * 0.035 ml
mass of oxygen gas = 0.040215 g
Number of moles of oxygen gas = 0.0402 g/(32 g/mol)
Number of moles of oxygen gas = 0.00125 moles
<em>Using the ideal gas equation, PV=nRT</em>
where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273)K = 310 K
<em>V = nRT/P</em>
V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm
V = 0.0318L or 31.8mL