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Studentka2010 [4]
3 years ago
7

Osmotic pressure Π is given by the relation:Π = iMRTwhere i is the van’t Hoff factor, M is the concentration of solute, R is the

gas constant, and T is the temperature. The osmotic pressure of sea water is approximately 24 atm at 25°C. What is the approximate concentration of salt in sea water (approximated by NaCl with i = 2)? (Note: Use R = 0.08 L•atm/mol•K.)
Chemistry
1 answer:
lions [1.4K]3 years ago
5 0

<u>Answer:</u> The concentration of solute is 0.503 mol/L

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=icRT

where,

\pi = osmotic pressure of the solution = 24 atm

i = Van't hoff factor = 2 (for NaCl)

c = concentration of solute = ?

R = Gas constant = 0.08\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

24atm=2\times c\times 0.08\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=0.503mol/L

Hence, the concentration of solute is 0.503 mol/L

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Given any two elements within a group, is the element with the larger atomic number likely to have a larger or smaller atomic ra
zhenek [66]

Answer:

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

Consider the example of sodium and potassium.

Sodium is present above the potassium with in same group i.e, group one.

The atomic number of sodium is 11 and potassium 19.

So potassium will have larger atomic radius as compared to sodium.

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

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3 years ago
What is the limiting reactant if 4.0 g of Nh3 react with 8.0 g of oxygen?
Mazyrski [523]
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8 0
3 years ago
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Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

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Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

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-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

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7 0
3 years ago
sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction
prisoha [69]

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

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2. Calculate K

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An acidic compound.

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