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Zanzabum
3 years ago
15

If the absorbance of the 10μM solution is 0.24, what should be the absorbance of the 50μM? Write answer to two decimal places.

Chemistry
1 answer:
Phoenix [80]3 years ago
3 0
Absorbance/Concentration = Constant.

That is, absorbance is directly proportional to concentration.

Therefore,

0.24/10 = x/50, where x is the absorbance at 50 micro m.

Calculating for x;
x=50* (0.24/10) = 1.20 (correct to 2 decimal places)
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5 0
3 years ago
What is the volume of 14.0g of nitrogen gas at STP?
lozanna [386]

Answer:

  • <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

  •   Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

  •    After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1)  of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>

  • n = mass in grams / molar mass
  • Atomic mass of nitrogen: 14.0 g/mol
  • Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
  • n = 14.0 g / 28.0 g/mol = 0.500 mol

<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>

  • 22.4 liter / mol = x  / 0.500 mol
  • Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

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5 0
3 years ago
Hello can somone help me ?
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