The answer is N2 + 3H2 yields 2NH3. The oxidation-reduction reaction means that there is electrons transfer during the reaction which means that the valence changed.
The mass number of this particular element is 37
An element with 17 protons will ALWAYS be chlorine.
The mass number is found by adding the element’s protons and neutrons.
protons+neutrons=mass number
Here is the math if needed,
17+20=37
Therefore the mass number is 37
I hope this helped!
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = 
so 1.54 moles of sodium azide will give =
mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = 