Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Answer:
Gas, Liquid, Solid.
Explanation:
please mark brianliest :)
a) The E might belong to group 13.
As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.
Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.
The Group 13 is the boron group which has the following elements:
- Boron
- Aluminium
- Gallium
- Indium
- Thallium
All these elements have the valency of +3.
To know more about Valency, refer to this link:
brainly.com/question/12717954
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Answer:
Explanation:
(a) melting point of S,S-hydrobenzoin should be 140°C. Since S,S-hydrobezoin and R,R-hydrobenzoin are enantiomeric pair so that their melting point and boilong point should be same.
(b) different melting point,
Racemic mixture must have different melting point in comparison to the pure enantiomers.
considering a racemic mixture, individual enantiometer possesses a greater affinity for its kind of molecules than for those of the other enantiometer
