Use PV=nRT to solve the equation. You need to solve for n (number of moles). Don’t forget to convert the temperature to kelvins by adding 25+273. Use 0.082057 for R.
No more solute will dissolve at that temperature, the temperature would have to be increased in order for more solute to dissolve.
We are given with the mass of pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.
Answer:
The correct answer is option A, that is, one valence electron in its third energy shell and option C, that is, 11 electrons and 11 protons.
Explanation:
The outermost electrons and the ones that take part in the process of bonding are termed as valence electrons. The atomic number of sodium is 11, thus, it possesses 11 protons and the atoms are neutral so it suggests that sodium has 11 electrons. By electronic configuration, it can be seen that in sodium, two electrons are present in the first shell, 8 in the second, and only one electron in the third shell, that is, 2.8.1. The electron present in the third shell is the valence electron.
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.