Answer:
Ek = (RT/zF)*ln ( [k+]o/[K+]i )
Explanation:
R = gas constant (8.31 J/Kmol)
T = Temperature (k)
F = Faraday constant (9.65 * 10exp4 coulomb/mole)
z = valence of the ion (1)
[k+]o = Extracellular K concentration in mM
[K+]i = Intracellular K concentration in mM
ln = logarithm with base e
Answer:
d. directly proportional to the number of moles of the gas.
Explanation:
The formula for the partial pressure of a gas is written as -
P = n * P ( total )
P = partial pressure of the gas ,
n = mole fraction of the gas ,
P ( total ) = Total pressure of the mixture of gas .
Hence , from the above equation ,
the partial pressure is directly proportional to the mole fraction , i.e. , on increasing the number of moles of gas , the partial pressure increases .
For Hydrocarbon combustion:
CₓHₐ + O₂ → xCO₂ + a/2 H₂O
Moles of CO₂ = 16.2 / 44 = 0.37
Moles of Carbon = 0.37
Moles of H₂O = 4.976 / 18 = 0.28
Moles of Hydrogen = 0.28 x 2 = 0.56
Molar ratio = C : H = 1 : 1.5
= 2 : 3
C₂H₃
Mass of empirical unit = 12 x 2 + 1 x 3
= 27
Mr = 54.09
Number of empirical units repeated: 54.09 / 27
= 2
Molecular formula = C₄H₆
Answer:
126 g of H₂O
Explanation:
Firstly, we complete the equation:
2H₂ + O₂ → 2H₂O
If we have 7 moles of hydrogen, we assume, the oxygen is in excess.
Ratio is 2:2.
2 moles of hydrogen can produce 2 moles of water
Then, 7 moles of H₂ must produce 7 moles of water.
We convert moles to mass → 7 mol . 18 g/mol = 126 g