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docker41 [41]
3 years ago
7

Imagine a cell with a semi-permeable membrane that is selective to K+ ions only. The internal solution contains 100 mM KCl and t

he external solution contains 1 mM KCl. What equation would you use to determine the potential (EK) developed across the membrane?
Chemistry
1 answer:
Scrat [10]3 years ago
7 0

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

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Does a hypothesis explains what the scientist thinks will happen during the experiment.
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Hypothesis is your opinion/ your thinking and yes, scientists have hypothesis before they do an experiment

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3 0
3 years ago
A 47.1 g sample of a metal is heated to 99.0°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at
wlad13 [49]

Answer : The metal used was iron (the specific heat capacity is 0.44J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of metal = ?

C_1 = specific heat of water = 4.18J/g^oC

m_1 = mass of metal = 47.1 g

m_2 = mass of water = 120 g

T_f = final temperature of water = 24.5^oC

T_1 = initial temperature of metal = 99^oC

T_2 = initial temperature of water = 21.4^oC

Now put all the given values in the above formula, we get

47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC

c_1=0.44J/g^oC

Form the value of specific heat of metal, we conclude that the metal used in this was iron.

Therefore, the metal used was iron (the specific heat capacity is 0.44J/g^oC).

3 0
3 years ago
Stings to the touch
likoan [24]
It’s A because bleach doesn’t sting on touch and you can’t taste it
8 0
3 years ago
Help!!!
Vedmedyk [2.9K]

Answer:

3.213J/mol

Explanation:

first specific heat capacity of water = 4200J/Kg K

Q=mass×temperature difference×specific heat capacity

Q=1.53×(24.3-20)×4200

Q=25.704 J

heat of solution= quantity of heat ÷ amount of substance

heat of solution= 25.704÷8

heat of solution= 3.213J/mol

8 0
3 years ago
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