<span>Mg + O2 > MgO. In reactant side, 2 O atoms and 1 Mg are present. In product side, 1 Mg and O atoms are present. Put 2 in product side to balance O atoms and 2 at Mg in reactant side to balance Mg atoms. Therefore the balanced equation becomes, 2Mg + O2 ----> 2MgO. Hope it helps.</span>
Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
Answer:
2.01V ( To three significant digits)
Explanation:
First we show the standard reduction potentials of Cu2+(aq)/Cu(s) system and Al3+(aq)/Al(s) system. We can clearly see from the balanced redox reaction equation that aluminium is the anode and was the oxidized specie while copper is the cathode and was the reduced specie. This observation is necessary when substituting values of concentration into the Nernst equation.
The next thing to do is to obtain the standard cell potential as shown in the image attached and subsequently substitute values of concentration and standard cell potential into the Nernst equation as shown. This gives the cell potential under the given conditions.
Answer:
12 grams of hydrogen gas
and 56 grams of nitrogen gas
The molar mass of ammonia is 17 g/mol.
68 grams of ammonia corresponds to
17g/mol
68g
=4moles
4 moles of ammonia will be obtained from
2
4×1
=2 moles of nitrogen and
2
4×3
=6 moles of hydrogen.
The molar masses of nitrogen and hydrogen are 28 g/mol and 2 g/mol respectively.
2 moles of nitrogen corresponds to 2×28=56 grams.
6 moles of hydrogen corresponds to 6×2=12 grams.
The concentration of NaOH is 25.00% by mass, it means that 25.00% of the mass of the solution is of NaOH. Hence:
