Chemistry, 29.03.2020 04:06 zoeedadoll

A solution of carbon tetrachloride and carbon disulfide containing 50% by weight of each is to be continuously distilled at a rate of 5000 kg/h.
(a) Determine the concentration of the mixture in terms of mole fractions. (b) Determine the average molecular weight of the mixture.
(c) Calculate the feed rate in kmol/h.

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Answer from: charismameeks

(a) CS₂ = 0.331, CCl₄ = 0.669; (b) 102 g/mol; (c) 49.1 kmol/h

Explanation:

Assume 5000 kg of mixture.

(a) Mole fractions

(i) Mass of each compound

Mass of CCl₄ = 2500 kg

Mass of CS₂ = 2500 kg

(ii) Moles of each compound

$\text{Moles of CCl}_{4} = \text{2500 kg CCl}_{4} \times \dfrac{\text{1 kmol CCl}_{4}}{\text{153.82 kg CCl}_{4}} = \text{16.3 kmol CCl}_{4}\\\\\text{Moles of CS}_{2} = \text{2500 kg CS}_{2} \times \dfrac{\text{1 kmol CS}_{2}}{\text{76.14 kg CS}_{2}} = \text{32.8 kmol CS}_{2}$

(iii) Mole fractions

Total moles = 16.3 kmol + 32.8 kmol = 49.1 kmol

The major component is usually called the solvent — Component 1 —carbon disulfide.

$\chi_{2} = \dfrac{n_{2}}{n_{\text{tot}}} =\dfrac{\text{16.3}}{\text{49.1}} = \mathbf{0.331}\\\\\chi_{1} = \dfrac{n_{1}}{n_{\text{tot}}} = \dfrac{\text{35.8}}{\text{49.1}} = \mathbf{0.669}$

The average molar mass is the weighted molar mass of each component.

$\text{MM} =\chi_{2} \text{M}_{2} + \chi_{1} \text{M}_{1} = 0.331 \times \dfrac{\text{153.82 g }}{\text{1 mol}} + 0.669 \times \dfrac{\text{76.14 g}}{\text{1 mol}} = \dfrac{\text{50.9 g}}{\text{1 mol }} + \dfrac{\text{50.9 g}}{\text{1 mol }}\\\\=\textbf{102 g/mol}$

(d) Feed rate

$\text{Feed rate} = \dfrac{\text{total moles}}{\text{time}} = \textbf{49.1 kmol/h}$

Answer from: Quest

answer: h+ its less than oh

explanation:

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.