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Jlenok [28]
3 years ago
9

What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac

ellular concentration of na ions is 27 mm at 20°c?
Chemistry
1 answer:
Ganezh [65]3 years ago
4 0

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions

• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

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umka21 [38]

Answer:

Defective collagen proteins affect an individuals joints, as <u>collagen is an important component protein in body</u><u> connective tissues</u>

Explanation:

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6 0
3 years ago
The compound ncl3 is nitrogen trichloride, but alcl3 is simply aluminum chloride. why
AveGali [126]
The answer to this question would be:
NCl3 is a molecular compound (two or more nonmetals), and therefore in its name prefixes indicate the number of each type of atom. so NCl3 is nitrogen trichloride<span>.
</span><span>The compound AlCl3 is an ionic compound (metal and nonmetal), and therefore does not require prefixes. so AlCl3 is aluminum chloride.
</span><span>
Both of nitrogen and chlorine is nonmetal, but aluminum is metal. Metal with nonmetal will make an ionic compound that doesn't need prefixes.</span>
5 0
3 years ago
Read 2 more answers
Pizza is an example of a homogeneous mixture.
Kay [80]
No its a example of heterogeneous mixture
4 0
3 years ago
What is the density of an<br> object that has a mass of<br> 120 g and volume of 5 ml?
Viktor [21]

The density of an  object ρ = 24 g/ml

<h3>Further explanation</h3>

Given

mass of an object = 120 g

volume = 5 ml

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Input the value :

ρ = 120 g : 5 ml

ρ = 24 g/ml

3 0
3 years ago
If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac
STALIN [3.7K]

<u>Answer:</u> The theoretical yield of barium sulfate is 50.9 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol

  • <u>For barium chloride:</u>

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = \frac{1}{1}\times 0.228=0.228mol of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = \frac{1}{1}\times 0.228=0.228moles of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g

Hence, the theoretical yield of barium sulfate is 50.9 grams

7 0
3 years ago
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