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dolphi86 [110]
3 years ago
11

Calculate the molar mass of a gas

Chemistry
1 answer:
spin [16.1K]3 years ago
4 0

Answer:

MOLAR MASS = 32 g/mol

Explanation:

Condition of standard temperature and pressure(STP) are as follow:

Temperature = 273 K

Pressure = 1 atm (or 100000 Pa)

Here atm is atmosphere and Pa is Pascal

STP conditions arte used for measuring gas density and volume using Ideal Gas Law.Here 1 mole of ideal gas occupies 22.4 L of volume.

According toi Ideal Gas Equation :

PV = nRT

where P = pressure, n= number of moles, V = volume ,R= Ideal Gas Constant and T= temperature

n=\frac{PV}{RT}

From question:

V=280 ml = 0.28 L

P = 1 atm

R=0.08205 L atm/K mol

T=273 K

Putting values in above formula :

n=\frac{1\times .280}{0.08205\times 273}

n = 0.0125 moles

Now n=\frac{given\ mass}{Molar\ mass}

Molar\ mass=\frac{given\ mass}{n}

given mass = 0.4 g (given)

Molar\ mass=\frac{0.4}{0.0125}

On solving we get:

Molar mass = 32 g/mol

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Answer: 9.09 %

Explanation:

To calculate the  percentage concentration by volume, we use the formula:

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Volume of ethanol (solute) = 30 ml

Volume of water (solvent) = 300 ml

Volume of solution= volume of solute + volume of solution = 30+ 300 = 330 ml

Putting values in above equation, we get:

\text{Volume percent of solution}=\frac{30}{30+300}\times 100=9.09\%

Hence, the volume percent of solution will be 9.09 %.

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Gaseous chlorine dioxide (ClO2) is used in bleaching flour and municipal water treatment in
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Taking into account the ideal gas law, the pressure is 2.52 atm.

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas are related by a simple formula called the ideal gas law. This equation relates the three variables if the amount of substance, number of moles n, remains constant. The universal constant of ideal gases R has the same value for all gaseous substances. The numerical value of R will depend on the units in which the other properties are worked.

P×V = n×R×T

In this case, you know:

  • P=?
  • V= 500 L
  • n= 52.1 moles
  • R= 0.082\frac{atmL}{molK}
  • T= 22 C= 295 K (being 0 C=273 K)

Replacing in the ideal gas law:

P×500 L = 52.1 moles ×0.082 \frac{atmL}{molK} ×295 K

Solving:

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2 years ago
Please help!! ASAP Calculate the pH of the solution after the addition of the following amounts of 0.0574 M HNO3 to a 80.0 ml so
alexira [117]

Answer:

a) 10.457.

b) 9.32.

c) 8.04.

d) 6.58.

e) 4.76.

f) 2.87.

Explanation:

  • Aziridine is an organic compounds containing the aziridine functional group, a three-membered heterocycle with one amine group (-NH-) and two methylene bridges (-CH2-). The parent compound is aziridine (or ethylene imine), with molecular formula C2H5N.
  • Aziridine has a basic character.
  • It has pKa = 8.04.
  • So, pKb = 14 – 8.04 = 5.96
  • Kb = 1.1 x 10⁻⁶.
  • If we denote Aziridine the symbol (Az),  It is dissociated in water as:

Az + H₂O → AzH⁺ + OH⁻

<u><em>a) 0.00 ml of HNO₃: </em></u>

There is only Az,

[OH⁻] = √(Kb.C)

Kb = 1.1 x 10⁻⁶. & C = 0.0750 M.

[OH⁻] = √(1.1 x 10⁻⁶)(0.075) = 2.867 x 10⁻⁴.

∵ pOH = - log[OH-] = - log (2.867 x 10⁻⁴) = 3.542.

∴ pH = 14 – pOH = 14 – 3.542 = 10.457.

<u><em>b) 5.27 ml of HNO₃</em></u>

  • To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
  • No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
  • No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol.
  • The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (0.302).
  • This will form a basic buffer in the presence of weak base (Az).

<em>pOH = pKb + log[salt]/[base] </em>

  • [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (0.302) / (85.27) = 3.54 x 10⁻³ M.
  • [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 0.302 mmol) / (85.27 ml) = 0.0668 M.
  • pOH = pKb + log[salt]/[base] = 5.96 + log[3.54 x 10⁻³]/[ 0.0668 M] = 4.68.
  • <em>pH = 14 – pOH = 14 – 4.68 = 9.32. </em>

<em />

<em><u>c) Volume of HNO₃ equal to half the equivalence point volume :</u></em>

  • At half equivalence point, the concentration of the salt formed is equal to the concentration of the remaining base (aziridine), [salt] = [base].
  • pOH = pKb + log[salt]/[base] = 5.96 + log[1.0] = 5.96.
  • pH = 14 – pOH = 14 – 5.96 = 8.04.

<u><em>d) 101 ml of HNO₃: </em></u>

  • To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
  • No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
  • No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (101.0 ml) = 5.7974 mmol.
  • The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (5.7974).
  • This will form a basic buffer in the presence of weak base (Az).

<em>pOH = pKb + log[salt]/[base] </em>

  • [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (5.7974) / (181.0) = 0.032 M.
  • [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 5.7974 mmol) / (181.0 ml) = 0.00112 M.
  • pOH = pKb + log[salt]/[base] = 5.96 + log[0.032]/[ 0.00112] = 7.416.
  • pH = 14 – pOH = 14 – 7.416 = 6.58.

<u><em>e) Volume of HNO₃ equal to the equivalence point :</em></u>

  • At the equivalence point the no. of millimoles of the base is equal to that of the acid.
  • Volume of HNO₃ needed for the equivalence point = (6.00 mmol) / (0.0574 mmol/ml) = 104.5 ml  

At the equivalence point:  

  • [Az] = 0.
  • [AzH⁺] = (6.00 mmol) / (80.0 + 104.5 ml) = 0.0325 M.
  • As Ka is very small, the dissociation of AzH⁺ can be negligible.  

Hence, [AzH⁺] at eqm ≈ 0.0325 M.

  • [H+] = √(Ka.C) = √(10⁻⁸˙⁰⁴ x 0.0325) = 1.72 x 10⁻⁵.
  • pH = - log[H+] = - log(1.72 x 10⁻⁵) = 4.76.

<u><em>f) 109 ml of HNO₃: </em></u>

  • No. of milli-moles of H⁺ added from HNO₃ = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol.
  • Which is higher than the no. of millimoles of the base (Az) = 6.0 mmol.
  • After the addition, [H⁺] = (6.257 - 6.00) / (80.0 + 109 mL) = 0.00136 M.
  • As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible.  
  • pH = -log[H⁺] = -log(0.00136) = 2.87.
3 0
3 years ago
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