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AleksandrR [38]
3 years ago
12

Calculate the mass of 45.0 L of NH3 at 57.0° C and 900. mm Hg.

Chemistry
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

Mass = 33.515 g

Explanation:

Given data:

Volume of ammonia = 45.0 L

Temperature = 57.0°C (57 +273 = 330 K)

Pressure = 900 mmHg

Mass of ammonia = ?

Solution:

Formula:

PV = nRT

We will use R = 62.364 (L * mmHg)/(mol * K) because pressure is given in mmHg.

900 mmHg × 45.0 L = n × 62.364L * mmHg/mol * K × 330 K

40500 mmHg.L = n ×  20580.12L * mmHg/mol

n= 40500 mmHg.L/ 20580.12L * mmHg/mol

n = 1.9679 mol

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 1.9679 mol × 17.031 g/mol

Mass = 33.515 g

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value.

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DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
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n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
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Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

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'n' specifies  → The energy and average distance from the nucleus.

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