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LenaWriter [7]
3 years ago
10

Determine which change of state is being described in this passage

Chemistry
2 answers:
alexandr402 [8]3 years ago
7 0

Explanation:

gas to liquid

correct option

......................

Softa [21]3 years ago
3 0

Answer:

Its c

Explanation:

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What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac
Ganezh [65]

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

4 0
3 years ago
What is the concentration of hydrogen ions in a solution of KOH with a pOH of 1.72?
bezimeni [28]

Answer:

[h+]= 5.25e-13

Explanation:

pH= 14-1.72= 12.28

[h+]= 5.25e-13

3 0
3 years ago
Why must sample spots be above the solvent in paper chromatography
marishachu [46]
Because if they are submerged in the solvent, they would dissolve! This would prevent them from seperating and not allow you to actually record anything
6 0
3 years ago
The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
worty [1.4K]
<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

4 0
3 years ago
Please answer, please answer please answer please answer<br>my last 2 chemistry question. ​
scoundrel [369]

Answer:

1. filtration and evaporation

2. i) water is added to the sand and salt mixture

ii) then the mixture is filtrated and so the sand and the salt water was seperated

iii) the salt water is heated with the help of burner until the water gets evaporated

iv) after the water gets evaporated, the salt is remained in the container

3. observation:

  • on adding water to the mixture, the salt is completely dissolved in the water
  • when filtrated the sand is seperated from the salt water
  • now the salt water when heated with the burner until the evaporation, the water is evaporated
  • the salt is precipitated and remained in the container

4. cautions:

  • while using the burner, we should be cautious with fire
  • the container that is heated should be holded with the help of a cloth to avoid heat
7 0
3 years ago
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