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Wittaler [7]
3 years ago
6

A 60 ml solution contains 200 ppm of iron. How many milliliters of water must be added to dilute this solution to a final concen

tration of 30 ppm?
O a 34 mi
Ob.540 ml
O G. 340 mL
d. 640 mL
Chemistry
1 answer:
AfilCa [17]3 years ago
8 0

A girl cycles for 3.00 hrs at a speed of 40.0 km/h. What distance did she ... A train travels at a speed of 30.0 km/hr and travel a distance of 240 km.. How long did it take the ... If a car travels 400 m in 20 seconds how fast is it going? Givens ... is in m/s2. 1. A man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s2.

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If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

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400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution
ch4aika [34]

<u>Answer:</u> The molecular weight of protein is 1.14\times 10^2g/mol

<u>Explanation:</u>

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

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT

where,

\pi = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of protein = 400 mg = 0.4 g   (Conversion factor:  1 g = 1000 mg)

M_{solute} = molar mass of protein = ?

V_{solution} = Volume of solution = 5.00 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

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

Putting values in above equation, we get:

0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol

Hence, the molecular weight of protein is 1.14\times 10^2g/mol

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The exposed metal rusts is an example of a chemical change because rust is an example of a chemical change in objects for example bicycles, scooters, etc.

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