did it hurt when you fell from heaven ;)
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:
The molar mass for the solute is 180.8 g/mol
Explanation:
We apply the colligative property of freezing point depression.
The formula is T° freezing pure solvent - T° freezing solution = Kf . m
T° freezing pure solvent - T° freezing solution → ΔT
Kf = 1.86 °C/m
m is molality. We must determine with data given.
Molality are the moles of solute contained in 1 kg of solvent
0° - (-1.8°C) = 1.86 °C/m . m
1.8°C / 1.86 m/°C = m → 0.968 mol/kg
We multiply the molality by kg of solute, to determine the moles
100 g . 1kg/1000 g = 0.1kg → 0.968 mol/kg . 0.1 kg = 0.0968 moles
To find the molar mass we make g/ mol → 17.5 g / 0.0968 mol =
180.8 g/mol
Answer:
A). Trial 1, because the average rate of the reaction is lower.
Explanation:
The first sentence elaborates that 'trial 1 had a lower level of concentration of the reactant that eventually led the reaction's average rate to fall.'
A trial having a higher level of concentration of a specific reactant will cause the concentration's average rate to rise. As per the graph provided, trial 2's average reaction rate is higher which shows that its concentration is greater and more reactive. However, the opposite happens in trial 1 where the amount of reactant is lesser in the concentration that leads the concentration to possess a lower rate of reaction. Thus, <u>option A</u> is the correct answer.
Answer:
The sample of 50 mg decay to 10 mg in 377.7 time.
Explanation:
Given that:- The exponential decay model for the decay after t days as:-
Where,
is the concentration at time t
is the initial concentration
Given:
= 50 mg
= 10 mg
So,
<u>The sample of 50 mg decay to 10 mg in 377.7 time.</u>