Answer:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
Explanation:
The standard formation reaction is the synthesis of 1 mole of a substance from its elements in their most stables forms under standard conditions. The balanced chemical equation is:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
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.
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There are a couple of ways in which you can express the concentration of a solution, and here they are: gram per liter (g/L), molarity (M), parts per million (ppm.), and percents (%).
As you can see, only M appears in your answers, which means that the correct option should be (2) 3.5 M.
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