Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For 
:-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
<u>Final concentration of NaOH = 0.75 M</u>
it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.
Answer:
O2 is a covalent substance while NaCl is an ionic substance
Explanation:
In O2 molecule, the bond is between 2 oxygen atoms which are non - metals. Thus, this is a covalent bond since it involves 2 non metals.
Whereas, for the NaCl molecule, the bond is between a metal sodium (Na) and a non metal Chloride(Cl) and thus we can say this is an ionic bond.
Thus the difference is that O2 is a covalent substance while NaCl is an ionic substance.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles = 
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = 
= 0.13moles
Number of moles of NH₃ = 
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
