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statuscvo [17]
3 years ago
7

Carbon disulfide is an important industrial solvent. It is prepared by the reaction of coke with sulfur dioxide. 5 C (s) + 2 SO2

(g) --> CS2 (l) + 4 CO (g) How many moles of CS2 form when 2.79 mol C react?
Chemistry
1 answer:
lutik1710 [3]3 years ago
5 0
I got 0.558 moles of CS2
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Question 6: Iron metal and chlorine gas react to form iron (III) chloride: 2 Fe(s) + 3 Cl2 (g) → 2 FeCl3 (s)
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10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.

What ois Theoretical yield ?

The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.

The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.

  • As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃

Therefore ,

10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂

  • As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃

Therefore ,

10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃

Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.

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4 0
3 years ago
Complete this nuclear reaction by selecting which particle would go in the
kompoz [17]

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Missing particles: ^{0}_{1}e^{+} (a positron) and an electron neutrino \nu_{\rm e}.

The nuclear equation would be:

{\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}.

Explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by 1.

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron ^{0}_{1}e^{+} and an electron neutrino \nu_e would be released. On the other hand, in a beta-minus decay, a negatively-charged electron \rm ^{0}_{1}e^{-} and an electron antineutrino \overline{\nu}_e would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the \rm Ne nucleus on the left-hand side is 10, meaning that the nucleus has a charge of +10. On the other hand, the atomic number of the \rm F nucleus on the right-hand side shows that this nucleus carries a charge of only +9.

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of +1. That rules out the possibility of the combination of one negatively-charged electron \rm ^{0}_{1}e^{-} (with a charge of -1) and an electron antineutrino \overline{\nu}_e (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino \nu_e, which carries no electric charge.)

4 0
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